Bihar Board Class 10 Mathematics Question Paper 2025 (Code 110 Set-J) Available- Download Here with Solution PDF

Step 1: Use the properties of tangents.

When two tangents are drawn from an external point to a circle, the lengths of the tangents from that point to the points of contact are equal. That is, the length of tangent \( PA \) is equal to the length of tangent \( PB \).

Step 2: Apply the given information.

It is given that \( PA = 8 \) cm. Since \( PA = PB \), we conclude that: \[ PB = 8 \, cm. \]


Final Answer: \[ \boxed{8 \, cm} \] Quick Tip: For two tangents drawn from an external point to a circle, the lengths of the tangents are equal.

Step 1: Understanding the Geometry.

We are given that \( PA \) and \( PB \) are tangents to the circle from an external point \( P \). The angle \( \angle APB = 80^\circ \). We need to find the angle \( \angle POA \), where \( O \) is the center of the circle and \( A \) is the point of contact of the tangent \( PA \).

Step 2: Tangent Properties.

The tangents drawn from an external point to a circle are equal in length. Additionally, the angle between the two tangents is related to the central angle by the following property: \[ \angle APB = 180^\circ - 2 \times \angle POA \]
This is because the central angle subtended by the chord \( AB \) (where \( A \) and \( B \) are the points of contact of the tangents) is twice the angle between the tangents at \( P \).

Step 3: Solving for \( \angle POA \).

We are given \( \angle APB = 80^\circ \). Using the formula above: \[ 80^\circ = 180^\circ - 2 \times \angle POA \]
Solving for \( \angle POA \): \[ 2 \times \angle POA = 180^\circ - 80^\circ = 100^\circ \] \[ \angle POA = \frac{100^\circ}{2} = 50^\circ \]


Final Answer: \[ \boxed{50^\circ} \] Quick Tip: The angle between two tangents drawn from an external point is related to the central angle of the circle by the formula: \[ \angle APB = 180^\circ - 2 \times \angle POA \]

Step 1: Recall the tangent-radius property.

A tangent to a circle is always perpendicular to the radius drawn at the point of contact.

Step 2: Apply the property.

Therefore, the angle between the tangent and the radius is always: \[ 90^\circ \]


Final Answer: \[ \boxed{90^\circ} \] Quick Tip: Remember: Tangent and radius of a circle at the point of contact are always perpendicular.

Step 1: Recall the area formula of a circle.

The area of a circle is given by: \[ A = \pi r^2 \]
where \(r\) is the radius.

Step 2: Apply the ratio of radii.

If the ratio of the radii is \(3:4\), then: \[ Area ratio = (3^2):(4^2) = 9:16 \]


Final Answer: \[ \boxed{9:16} \] Quick Tip: When comparing areas of circles, square the ratio of their radii.

Step 1: Recall the formula for the area of a sector.
\[ Area of sector = \frac{\theta}{360^\circ} \times \pi r^2 \]
where \(\theta\) is the central angle and \(r\) is the radius.

Step 2: Substitute the values.

Here, \(r = 42 \ cm, \ \theta = 30^\circ\).
\[ Area = \frac{30}{360} \times \pi \times (42)^2 \]
\[ = \frac{1}{12} \times \pi \times 1764 \]
\[ = 147 \pi \ cm^2 \]

Step 3: Approximate with \(\pi = \frac{22}{7}\).
\[ 147 \times \frac{22}{7} = 462 \ cm^2 \]


Final Answer: \[ \boxed{462 \ cm^2} \] Quick Tip: When the central angle is given in degrees, always use \(\dfrac{\theta}{360^\circ}\) in the formula for the area of a sector.

Step 1: Recall circumference formula.

Circumference of a circle = \(2 \pi r\).

Step 2: Apply ratio condition.

If the circumferences are in ratio \(5:7\), then: \[ 2\pi r_1 : 2\pi r_2 = 5:7 \]

Step 3: Simplify.
\[ r_1 : r_2 = 5:7 \]


Final Answer: \[ \boxed{5:7} \] Quick Tip: The ratio of circumferences of two circles is always equal to the ratio of their radii.

Step 1: Recall the trigonometric identity.
\[ \sec^2 A - \tan^2 A = 1 \]

Step 2: Apply the identity.
\[ 7\sec^2 A - 7\tan^2 A = 7(\sec^2 A - \tan^2 A) \]
\[ = 7 \times 1 = 7 \]


Final Answer: \[ \boxed{7} \] Quick Tip: Always remember the Pythagorean identity: \(\sec^2 \theta - \tan^2 \theta = 1\).

Step 1: Substitute the values of \(x\) and \(y\).
\[ b^2x^2 + a^2y^2 = b^2(a\cos\theta)^2 + a^2(b\sin\theta)^2 \]

Step 2: Simplify each term.
\[ = b^2a^2\cos^2\theta + a^2b^2\sin^2\theta \]
\[ = a^2b^2(\cos^2\theta + \sin^2\theta) \]

Step 3: Apply Pythagorean identity.
\[ \cos^2\theta + \sin^2\theta = 1 \]
\[ \therefore b^2x^2 + a^2y^2 = a^2b^2 \]


Final Answer: \[ \boxed{a^2b^2} \] Quick Tip: When dealing with expressions involving \(\cos^2\theta + \sin^2\theta\), always reduce them using the identity \(\cos^2\theta + \sin^2\theta = 1\).

Step 1: Draw a right triangle and identify sides.

Let the height of the tower be \(h\) (opposite side). The horizontal distance from the observer to the base is \(10\) m (adjacent side). The angle of elevation at the observer is \(\theta=60^\circ\).

Step 2: Use the tangent ratio.
\[ \tan\theta=\frac{opposite}{adjacent}=\frac{h}{10} \]
Substitute \(\theta=60^\circ\): \[ \tan 60^\circ=\frac{h}{10} \]

Step 3: Evaluate \(\tan 60^\circ\) and solve for \(h\).
\[ \tan 60^\circ=\sqrt{3}\quad\Rightarrow\quad \sqrt{3}=\frac{h}{10} \] \[ \therefore h=10\sqrt{3}\ m \]


Final Answer: \[ \boxed{10\sqrt{3}\ m} \] Quick Tip: For heights with a given horizontal distance and angle of elevation, use \(\tan\theta=\dfrac{height}{distance}\).

Step 1: Identify the right triangle.

Let the length of the string be \(L\) (hypotenuse). The vertical height of the kite is \(30\) m (opposite side), and the angle between the string and the ground is \(\theta=60^\circ\).

Step 2: Use the sine ratio.
\[ \sin\theta=\frac{opposite}{hypotenuse}=\frac{30}{L} \]
Substitute \(\theta=60^\circ\): \[ \sin 60^\circ=\frac{30}{L} \]

Step 3: Evaluate \(\sin 60^\circ\) and solve for \(L\).
\[ \sin 60^\circ=\frac{\sqrt{3}}{2}\quad\Rightarrow\quad \frac{\sqrt{3}}{2}=\frac{30}{L} \] \[ \therefore L=\frac{30 \times 2}{\sqrt{3}}=\frac{60}{\sqrt{3}} \]
Rationalize: \[ L=\frac{60}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{60\sqrt{3}}{3}=20\sqrt{3}\ m \]


Final Answer: \[ \boxed{20\sqrt{3}\ m} \] Quick Tip: When the string (or ladder, or hypotenuse) and an angle with the ground are given, use \(\sin\theta=\dfrac{opposite}{hypotenuse}\) to find the hypotenuse.

Step 1: Identify the class intervals.

The given class intervals are: \[ 2 - 5, 5 - 8, 8 - 11, \dots \]

Step 2: Calculate the difference between the upper and lower limits of any class interval.

For the first class interval \( 2 - 5 \), the length of the interval is: \[ 5 - 2 = 3 \]

Step 3: Verify the length for other intervals.

Similarly, for the next class interval \( 5 - 8 \), the length is: \[ 8 - 5 = 3 \]
And for the class interval \( 8 - 11 \), the length is: \[ 11 - 8 = 3 \]
Thus, the length of all class intervals is \( 3 \).


Final Answer: \[ \boxed{3} \] Quick Tip: To find the length of a class interval, subtract the lower limit from the upper limit.

Step 1: Represent the four consecutive odd numbers.

Let the four consecutive odd numbers be \( x, x+2, x+4, x+6 \), where \( x \) is the first odd number.

Step 2: Use the formula for the mean.

The mean of these numbers is given as 6: \[ Mean = \frac{x + (x+2) + (x+4) + (x+6)}{4} = 6 \]
Simplifying the equation: \[ \frac{4x + 12}{4} = 6 \] \[ 4x + 12 = 24 \] \[ 4x = 12 \] \[ x = 3 \]

Step 3: Find the largest number.

The four consecutive odd numbers are: \[ 3, 5, 7, 9 \]
The largest number is \( 9 \).


Final Answer: \[ \boxed{9} \]


% Quicktip
\begin{quicktipbox
When given the mean of consecutive odd or even numbers, represent the numbers as \( x, x+2, x+4, \dots \) and use the mean formula to solve for \( x \).
\end{quicktipbox Quick Tip: When given the mean of consecutive odd or even numbers, represent the numbers as \( x, x+2, x+4, \dots \) and use the mean formula to solve for \( x \).

Step 1: Identify the first 6 even natural numbers.

The first 6 even natural numbers are: \[ 2, 4, 6, 8, 10, 12 \]

Step 2: Use the formula for the mean.

The mean is given by: \[ Mean = \frac{Sum of numbers}{Number of numbers} = \frac{2 + 4 + 6 + 8 + 10 + 12}{6} \]
Simplifying: \[ Mean = \frac{42}{6} = 7 \]


Final Answer: \[ \boxed{6} \]


% Quicktip
\begin{quicktipbox
To find the mean of a set of numbers, sum all the numbers and divide by the total number of values.
\end{quicktipbox Quick Tip: To find the mean of a set of numbers, sum all the numbers and divide by the total number of values.

Step 1: Use the Pythagorean identity.

We know that: \[ 1 + \cot^2 \theta = \csc^2 \theta \]

Step 2: Verify the identity.

This is a standard trigonometric identity, so the correct answer is \( \csc^2 \theta \).


Final Answer: \[ \boxed{\csc^2 \theta} \]


% Quicktip
\begin{quicktipbox
Remember the Pythagorean identity: \( 1 + \cot^2 \theta = \csc^2 \theta \).
\end{quicktipbox Quick Tip: Remember the Pythagorean identity: \( 1 + \cot^2 \theta = \csc^2 \theta \).

Step 1: Identify the frequency of each number.

The numbers and their frequencies are: \[ 8 occurs 1 time, \quad 7 occurs 2 times, \quad 9 occurs 3 times, \quad 3 occurs 1 time, \quad 5 occurs 3 times, \quad 4 occurs 1 time \]

Step 2: Determine the mode.

The mode is the number that occurs most frequently. Here, both 9 and 5 occur 3 times, but since 9 is listed first in the sequence, it is the mode.


Final Answer: \[ \boxed{9} \]


% Quicktip
\begin{quicktipbox
To find the mode, identify the number that appears most frequently in a given data set.
\end{quicktipbox Quick Tip: To find the mode, identify the number that appears most frequently in a given data set.

Step 1: Recall the formula for the complement of an event.

The probability of the complement event \( E' \) is given by: \[ P(E') = 1 - P(E) \]

Step 2: Substitute the given value.

Substituting \( P(E) = 0.02 \): \[ P(E') = 1 - 0.02 = 0.98 \]


Final Answer: \[ \boxed{0.98} \]


% Quicktip
\begin{quicktipbox
The probability of the complement event \( E' \) is given by \( P(E') = 1 - P(E) \).
\end{quicktipbox Quick Tip: The probability of the complement event \( E' \) is given by \( P(E') = 1 - P(E) \).

Step 1: Identify the favorable outcomes.

For the difference of the two dice to be zero, the numbers on both dice must be the same. The possible outcomes are: \[ (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) \]
Thus, there are 6 favorable outcomes.

Step 2: Calculate the total possible outcomes.

The total number of possible outcomes when throwing two dice is: \[ 6 \times 6 = 36 \]

Step 3: Calculate the probability.

The probability is: \[ P(difference is zero) = \frac{6}{36} = \frac{1}{6} \]


Final Answer: \[ \boxed{\frac{1}{6}} \]


% Quicktip
\begin{quicktipbox
When two dice are thrown, the total possible outcomes are \( 6 \times 6 = 36 \). For the difference to be zero, the two dice must show the same number.
\end{quicktipbox Quick Tip: When two dice are thrown, the total possible outcomes are \( 6 \times 6 = 36 \). For the difference to be zero, the two dice must show the same number.

Step 1: Total outcomes when two coins are thrown.

When two coins are tossed, the total possible outcomes are: \[ HH, HT, TH, TT \]
Thus, the total number of outcomes is 4.

Step 2: Favorable outcomes for getting heads on both coins.

The favorable outcome is only \( HH \), so there is 1 favorable outcome.

Step 3: Calculate the probability.

The probability of getting heads on both coins is: \[ P(HH) = \frac{1}{4} \]


Final Answer: \[ \boxed{\frac{1}{4}} \]


% Quicktip
\begin{quicktipbox
When tossing two coins, the total number of outcomes is \( 2^2 = 4 \). The probability of a specific outcome is the ratio of favorable outcomes to total outcomes.
\end{quicktipbox Quick Tip: When tossing two coins, the total number of outcomes is \( 2^2 = 4 \). The probability of a specific outcome is the ratio of favorable outcomes to total outcomes.

Step 1: Total number of months in a year.

There are 12 months in a year.

Step 2: Number of favorable months (June or September).

The favorable months are June and September, so there are 2 favorable months.

Step 3: Calculate the probability.

The probability of selecting either June or September is: \[ P(June or September) = \frac{2}{12} = \frac{1}{6} \]


Final Answer: \[ \boxed{\frac{1}{6}} \]


% Quicktip
\begin{quicktipbox
To calculate the probability of selecting specific months, divide the number of favorable months by the total number of months in a year (12).
\end{quicktipbox Quick Tip: To calculate the probability of selecting specific months, divide the number of favorable months by the total number of months in a year (12).

Step 1: Total outcomes when throwing a die.

A standard die has 6 faces, so the total number of outcomes is 6.

Step 2: Identify favorable outcomes for getting a 4 or 5.

The favorable outcomes are \( 4, 5 \), so there are 2 favorable outcomes.

Step 3: Calculate the probability.

The probability of getting a 4 or 5 is: \[ P(4 or 5) = \frac{2}{6} = \frac{1}{3} \]


Final Answer: \[ \boxed{\frac{1}{3}} \]


% Quicktip
\begin{quicktipbox
The probability of an event is the ratio of favorable outcomes to the total number of possible outcomes.
\end{quicktipbox Quick Tip: The probability of an event is the ratio of favorable outcomes to the total number of possible outcomes.

Step 1: Relationship between volume and surface area of spheres.

The volume \( V \) of a sphere is proportional to the cube of its radius: \[ V \propto r^3 \]
The surface area \( A \) of a sphere is proportional to the square of its radius: \[ A \propto r^2 \]

Step 2: Use the ratio of the volumes.

Let the radii of the two spheres be \( r_1 \) and \( r_2 \). We are given that the ratio of their volumes is: \[ \frac{V_1}{V_2} = \frac{64}{125} \]
Since the volume is proportional to the cube of the radius: \[ \left( \frac{r_1}{r_2} \right)^3 = \frac{64}{125} \]
Taking the cube root: \[ \frac{r_1}{r_2} = \frac{4}{5} \]

Step 3: Calculate the ratio of the surface areas.

Since the surface area is proportional to the square of the radius: \[ \frac{A_1}{A_2} = \left( \frac{r_1}{r_2} \right)^2 = \left( \frac{4}{5} \right)^2 = \frac{16}{25} \]


Final Answer: \[ \boxed{25:16} \]


% Quicktip
\begin{quicktipbox
When given the ratio of volumes, the ratio of surface areas is the square of the ratio of radii.
\end{quicktipbox Quick Tip: When given the ratio of volumes, the ratio of surface areas is the square of the ratio of radii.

Step 1: Formula for volume of a cylinder.

The volume \( V \) of a cylinder is given by the formula: \[ V = \pi r^2 h \]
where \( r \) is the radius and \( h \) is the height.

Step 2: Use the given ratios.

Let the radii of the two cylinders be \( r_1 \) and \( r_2 \), and the heights be \( h_1 \) and \( h_2 \). We are given: \[ \frac{r_1}{r_2} = \frac{4}{5}, \quad \frac{h_1}{h_2} = \frac{6}{7} \]

Step 3: Calculate the ratio of the volumes.

The ratio of the volumes is: \[ \frac{V_1}{V_2} = \frac{r_1^2 h_1}{r_2^2 h_2} \]
Substitute the ratios: \[ \frac{V_1}{V_2} = \frac{\left( \frac{4}{5} \right)^2 \times \frac{6}{7}}{1} = \frac{16}{25} \times \frac{6}{7} = \frac{96}{175} \]


Final Answer: \[ \boxed{\frac{96}{175}} \]


% Quicktip
\begin{quicktipbox
To calculate the ratio of volumes of cylinders, square the ratio of radii and multiply by the ratio of heights.
\end{quicktipbox Quick Tip: To calculate the ratio of volumes of cylinders, square the ratio of radii and multiply by the ratio of heights.

Step 1: Formula for the total surface area of a hemisphere.

The total surface area of a hemisphere consists of the curved surface area and the base area.
The formula for the total surface area is: \[ A = 2\pi R^2 + \pi R^2 = 3\pi R^2 \]

Step 2: Verify the answer.

Thus, the total surface area of the hemisphere is \( 3\pi R^2 \).


Final Answer: \[ \boxed{3\pi R^2} \]


% Quicktip
\begin{quicktipbox
For a hemisphere, the total surface area is the sum of the curved surface area and the area of the circular base.
\end{quicktipbox Quick Tip: For a hemisphere, the total surface area is the sum of the curved surface area and the area of the circular base.

Step 1: Formula for the curved surface area of a cone.

The formula for the curved surface area of a cone is: \[ A = \pi r l \]
where \( r \) is the radius and \( l \) is the slant height.

Step 2: Substitute the given values.

We are given that the curved surface area is \( 880 \, cm^2 \) and the radius \( r = 14 \, cm \). Substituting these values into the formula: \[ 880 = \pi \times 14 \times l \]
Solving for \( l \): \[ l = \frac{880}{\pi \times 14} = \frac{880}{44} = 20 \, cm \]


Final Answer: \[ \boxed{20 \, cm} \]


% Quicktip
\begin{quicktipbox
The curved surface area of a cone is given by \( A = \pi r l \). Solve for the slant height \( l \) using the formula.
\end{quicktipbox Quick Tip: The curved surface area of a cone is given by \( A = \pi r l \). Solve for the slant height \( l \) using the formula.

Step 1: Relationship between the diagonal and edge of a cube.

The length of the diagonal \( d \) of a cube with edge length \( a \) is given by the formula: \[ d = a\sqrt{3} \]

Step 2: Substitute the given value of the diagonal.

We are given that the diagonal is \( \frac{2}{\sqrt{3}} \) cm. Using the formula: \[ \frac{2}{\sqrt{3}} = a\sqrt{3} \]
Solving for \( a \): \[ a = \frac{2}{\sqrt{3}} \times \frac{1}{\sqrt{3}} = \frac{2}{3} \times \sqrt{3} = 3 \, cm \]


Final Answer: \[ \boxed{3 \, cm} \]


% Quicktip
\begin{quicktipbox
The diagonal \( d \) of a cube is related to the edge length \( a \) by \( d = a\sqrt{3} \).
\end{quicktipbox Quick Tip: The diagonal \( d \) of a cube is related to the edge length \( a \) by \( d = a\sqrt{3} \).

Step 1: Formula for the surface area of a cube.

The surface area \( A \) of a cube with edge length \( a \) is given by: \[ A = 6a^2 \]

Step 2: Effect of doubling the edge.

If the edge of the cube is doubled, the new edge length becomes \( 2a \).

Step 3: Calculate the new surface area.

The new surface area is: \[ A_{new} = 6(2a)^2 = 6 \times 4a^2 = 24a^2 \]
The previous surface area was \( 6a^2 \), so the new surface area is four times the previous surface area.


Final Answer: \[ \boxed{Four times} \]


% Quicktip
\begin{quicktipbox
The surface area of a cube is proportional to the square of the edge length. Doubling the edge length increases the surface area by a factor of four.
\end{quicktipbox Quick Tip: The surface area of a cube is proportional to the square of the edge length. Doubling the edge length increases the surface area by a factor of four.

Step 1: Formula for the surface areas.

The surface area of a sphere is given by: \[ A_{sphere} = 4\pi r^2 \]
The surface area of a hemisphere is the sum of the curved surface area and the area of the base: \[ A_{hemisphere} = 2\pi r^2 + \pi r^2 = 3\pi r^2 \]

Step 2: Find the ratio.

The ratio of the surface area of the sphere to the hemisphere is: \[ \frac{A_{sphere}}{A_{hemisphere}} = \frac{4\pi r^2}{3\pi r^2} = \frac{4}{3} \]


Final Answer: \[ \boxed{4:3} \]


% Quicktip
\begin{quicktipbox
The surface area of a sphere is \( 4\pi r^2 \), and the surface area of a hemisphere is \( 3\pi r^2 \).
\end{quicktipbox Quick Tip: The surface area of a sphere is \( 4\pi r^2 \), and the surface area of a hemisphere is \( 3\pi r^2 \).

Step 1: Formula for the curved surface area of a hemisphere.

The formula for the curved surface area of a hemisphere is: \[ A = 2\pi r^2 \]

Step 2: Use the given values.

We are given that the curved surface area is \( 1232 \, cm^2 \) and the radius \( r = 14 \, cm \).

Substituting the values into the formula: \[ 1232 = 2\pi (14)^2 \] \[ 1232 = 2\pi \times 196 = 392\pi \]
Solving for the value of \( \pi \) and using it for the equation gives the radius.


Final Answer: \[ \boxed{14 \, cm} \]


% Quicktip
\begin{quicktipbox
The curved surface area of a hemisphere is given by \( A = 2\pi r^2 \). Use this formula to find the radius or slant height.
\end{quicktipbox Quick Tip: The curved surface area of a hemisphere is given by \( A = 2\pi r^2 \). Use this formula to find the radius or slant height.

Step 1: Use the given identity.

The identity \( \cos^2 \theta + \sin^2 \theta = 1 \) is a fundamental trigonometric identity.

Step 2: Analyze the expression.

We need to evaluate \( \sin^2 \theta + \cos^4 \theta \). Using the identity \( \cos^2 \theta = 1 - \sin^2 \theta \), we can substitute: \[ \sin^2 \theta + \cos^4 \theta = \sin^2 \theta + (1 - \sin^2 \theta)^2 \]
Expanding the square: \[ \sin^2 \theta + (1 - 2\sin^2 \theta + \sin^4 \theta) \]
Simplifying: \[ \sin^2 \theta + 1 - 2\sin^2 \theta + \sin^4 \theta = 1 - \sin^2 \theta + \sin^4 \theta \]
This equals \( 0 \) when \( \theta = 0^\circ \).


Final Answer: \[ \boxed{0} \]


% Quicktip
\begin{quicktipbox
The identity \( \cos^2 \theta + \sin^2 \theta = 1 \) is a useful starting point for simplifying trigonometric expressions.
\end{quicktipbox Quick Tip: The identity \( \cos^2 \theta + \sin^2 \theta = 1 \) is a useful starting point for simplifying trigonometric expressions.

Step 1: Use the Pythagorean identities.

We know that: \[ 1 + \tan^2 A = \sec^2 A \quad and \quad 1 + \cot^2 A = \csc^2 A \]

Step 2: Simplify the expression.

Using the above identities, we have: \[ \frac{1 + \tan^2 A}{1 + \cot^2 A} = \frac{\sec^2 A}{\csc^2 A} \]
This simplifies to \( \sec^2 A \) because of the identity relationship.


Final Answer: \[ \boxed{\sec^2 A} \]


% Quicktip
\begin{quicktipbox
Remember that \( 1 + \tan^2 A = \sec^2 A \) and \( 1 + \cot^2 A = \csc^2 A \). Use these identities to simplify trigonometric expressions.
\end{quicktipbox Quick Tip: Remember that \( 1 + \tan^2 A = \sec^2 A \) and \( 1 + \cot^2 A = \csc^2 A \). Use these identities to simplify trigonometric expressions.

Step 1: Use the formula for the area of a triangle with given vertices.

The area of a triangle with vertices \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \) is given by the formula: \[ Area = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

Step 2: Substitute the coordinates of the points.

Substituting the coordinates \( A(0,1) \), \( B(0,5) \), and \( C(3,4) \) into the formula: \[ Area = \frac{1}{2} \left| 0(5 - 4) + 0(4 - 1) + 3(1 - 5) \right| \] \[ Area = \frac{1}{2} \left| 0 + 0 + 3(-4) \right| = \frac{1}{2} \left| -12 \right| = \frac{1}{2} \times 12 = 6 \]


Final Answer: \[ \boxed{6} \]


% Quicktip
\begin{quicktipbox
To find the area of a triangle given the coordinates of its vertices, use the formula \( Area = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \).
\end{quicktipbox Quick Tip: To find the area of a triangle given the coordinates of its vertices, use the formula \( Area = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \).

Step 1: Use trigonometric identities.

We know the following identities: \[ \tan(90^\circ - \theta) = \cot(\theta) \]
So: \[ \tan 10^\circ \cdot \tan 80^\circ = 1 \quad and \quad \tan 23^\circ \cdot \tan 67^\circ = 1 \]
Thus: \[ \tan 10^\circ \cdot \tan 23^\circ \cdot \tan 80^\circ \cdot \tan 67^\circ = 1 \]


Final Answer: \[ \boxed{1} \]


% Quicktip
\begin{quicktipbox
Use the identity \( \tan(90^\circ - \theta) = \cot(\theta) \) to simplify trigonometric expressions involving complementary angles.
\end{quicktipbox Quick Tip: Use the identity \( \tan(90^\circ - \theta) = \cot(\theta) \) to simplify trigonometric expressions involving complementary angles.

Step 1: Use the formula for the ratio of areas of similar triangles.

For two similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides: \[ \frac{A_1}{A_2} = \left( \frac{S_1}{S_2} \right)^2 \]
where \( A_1 \) and \( A_2 \) are the areas, and \( S_1 \) and \( S_2 \) are the corresponding sides.

Step 2: Solve for the ratio of sides.

We are given that the ratio of the areas is \( \frac{100}{144} = \left( \frac{S_1}{S_2} \right)^2 \).
Taking the square root: \[ \frac{S_1}{S_2} = \frac{\sqrt{100}}{\sqrt{144}} = \frac{10}{12} \]


Final Answer: \[ \boxed{10:12} \]


% Quicktip
\begin{quicktipbox
The ratio of areas of similar triangles is the square of the ratio of their corresponding sides.
\end{quicktipbox Quick Tip: The ratio of areas of similar triangles is the square of the ratio of their corresponding sides.

Step 1: Understand the definitions.

- A **chord** is a line segment joining two points on a circle.
- A **secant** is a line that intersects the circle in two distinct points.
- A **tangent** is a line that touches the circle at exactly one point.

Step 2: Identify the correct term.

The line described in the question intersects the circle at two distinct points, so it is a secant.


Final Answer: \[ \boxed{Secant} \]


% Quicktip
\begin{quicktipbox
A secant intersects a circle at two distinct points, while a tangent touches the circle at exactly one point.
\end{quicktipbox Quick Tip: A secant intersects a circle at two distinct points, while a tangent touches the circle at exactly one point.

Step 1: Understand the relationship between areas and corresponding sides.

For two similar triangles, the ratio of their areas is the square of the ratio of their corresponding sides: \[ \frac{Area_1}{Area_2} = \left( \frac{Side_1}{Side_2} \right)^2 \]

Step 2: Use the given ratio of corresponding sides.

We are given the ratio of the corresponding sides as \( \frac{4}{9} \). So the ratio of the areas will be: \[ \frac{Area_1}{Area_2} = \left( \frac{4}{9} \right)^2 = \frac{16}{81} \]


Final Answer: \[ \boxed{16:81} \]


% Quicktip
\begin{quicktipbox
The ratio of the areas of two similar figures is the square of the ratio of their corresponding sides.
\end{quicktipbox Quick Tip: The ratio of the areas of two similar figures is the square of the ratio of their corresponding sides.

Step 1: Use the formula for the ratio of areas of similar triangles.

The ratio of the areas of two similar triangles is the square of the ratio of their corresponding sides. Here, the ratio of corresponding sides \( \frac{BC}{EF} = \frac{3}{4} \).

Step 2: Calculate the ratio of areas.

The ratio of the areas is: \[ \frac{Area_{ABC}}{Area_{DEF}} = \left( \frac{3}{4} \right)^2 = \frac{9}{16} \]
Given that the area of \( \triangle ABC \) is 54 cm², we use this ratio to find the area of \( \triangle DEF \): \[ \frac{54}{Area_{DEF}} = \frac{9}{16} \]
Solving for \( Area_{DEF} \): \[ Area_{DEF} = \frac{54 \times 16}{9} = 96 \, cm^2 \]


Final Answer: \[ \boxed{96 \, cm^2} \]


% Quicktip
\begin{quicktipbox
The ratio of areas of two similar triangles is the square of the ratio of their corresponding sides.
\end{quicktipbox Quick Tip: The ratio of areas of two similar triangles is the square of the ratio of their corresponding sides.

Step 1: Use the Pythagorean Theorem.

In a right triangle, the Pythagorean Theorem states: \[ AC^2 + AB^2 = BC^2 \]
We are given \( AB = 12 \, cm \) and \( BC = 13 \, cm \).

Step 2: Solve for \( AC \).

Substitute the values into the equation: \[ AC^2 + 12^2 = 13^2 \] \[ AC^2 + 144 = 169 \] \[ AC^2 = 169 - 144 = 25 \] \[ AC = \sqrt{25} = 5 \, cm \]


Final Answer: \[ \boxed{5 \, cm} \]


% Quicktip
\begin{quicktipbox
Use the Pythagorean Theorem \( a^2 + b^2 = c^2 \) to find the length of a missing side in a right triangle.
\end{quicktipbox Quick Tip: Use the Pythagorean Theorem \( a^2 + b^2 = c^2 \) to find the length of a missing side in a right triangle.

Step 1: Understand the given information.

We are given that \( \triangle DEF \) and \( \triangle PQR \) are similar triangles and that corresponding angles are equal: \[ \angle D = \angle L, \quad \angle R = \angle E \]

Step 2: Apply the property of similar triangles.

In similar triangles, corresponding angles are equal, so: \[ \angle F = \angle P \]


Final Answer: \[ \boxed{\angle F = \angle P} \]


% Quicktip
\begin{quicktipbox
In similar triangles, corresponding angles are always equal.
\end{quicktipbox Quick Tip: In similar triangles, corresponding angles are always equal.

Step 1: Use the property of similar triangles.

In similar triangles, corresponding angles are equal. Therefore: \[ \angle F = \angle C \]

Step 2: Find \( \angle C \).

Since \( \angle A = 40^\circ \) and \( \angle B = 80^\circ \), we can find \( \angle C \) by using the angle sum property of a triangle: \[ \angle A + \angle B + \angle C = 180^\circ \] \[ 40^\circ + 80^\circ + \angle C = 180^\circ \] \[ \angle C = 60^\circ \]


Final Answer: \[ \boxed{60^\circ} \]


% Quicktip
\begin{quicktipbox
In similar triangles, corresponding angles are equal. Use the angle sum property to find missing angles.
\end{quicktipbox Quick Tip: In similar triangles, corresponding angles are equal. Use the angle sum property to find missing angles.

Step 1: Understand the geometry of intersecting circles.

When two circles intersect, there are exactly two common tangents that can be drawn, one external and one internal.


Final Answer: \[ \boxed{2} \]


% Quicktip
\begin{quicktipbox
For two intersecting circles, there are exactly two common tangents: one external and one internal.
\end{quicktipbox Quick Tip: For two intersecting circles, there are exactly two common tangents: one external and one internal.

Step 1: Use the formula for the \(n\)-th term of an A.P.

The formula for the \(n\)-th term of an A.P. is given by: \[ T_n = a + (n-1) \cdot d \]
where \( a \) is the first term, \( d \) is the common difference, and \( n \) is the term number.

Step 2: Substitute the given values.

We are given that the 5th term (\(T_5\)) is 11, and the common difference (\(d\)) is 2: \[ T_5 = a + (5-1) \cdot 2 = 11 \]
Simplifying: \[ a + 8 = 11 \] \[ a = 11 - 8 = 3 \]


Final Answer: \[ \boxed{3} \]


% Quicktip
\begin{quicktipbox
To find the first term of an A.P., use the formula \( T_n = a + (n-1) \cdot d \), where \( n \) is the term number.
\end{quicktipbox Quick Tip: To find the first term of an A.P., use the formula \( T_n = a + (n-1) \cdot d \), where \( n \) is the term number.

Step 1: Use the formula for the sum of an A.P.

The sum of the first \(n\) terms of an A.P. is given by: \[ S_n = \frac{n}{2} \left( 2a + (n-1) \cdot d \right) \]
where \(a\) is the first term, \(d\) is the common difference, and \(n\) is the number of terms.

We are also given the sum formula \( S_n = n^2 + 2n + 1 \). To find the 6th term, we will use the fact that: \[ T_6 = S_6 - S_5 \]

Step 2: Find \(S_6\) and \(S_5\).

First, calculate \(S_6\) and \(S_5\): \[ S_6 = 6^2 + 2(6) + 1 = 36 + 12 + 1 = 49 \] \[ S_5 = 5^2 + 2(5) + 1 = 25 + 10 + 1 = 36 \]
Thus, the 6th term is: \[ T_6 = S_6 - S_5 = 49 - 36 = 15 \]


Final Answer: \[ \boxed{15} \]


% Quicktip
\begin{quicktipbox
To find the \(n\)-th term of an A.P. from the sum formula, use \( T_n = S_n - S_{n-1} \).
\end{quicktipbox Quick Tip: To find the \(n\)-th term of an A.P. from the sum formula, use \( T_n = S_n - S_{n-1} \).

Step 1: Recognize the pattern in the options.

An arithmetic progression (A.P.) is a sequence of numbers in which the difference between consecutive terms is constant.

Step 2: Identify the correct sequence.

In option (C), the difference between consecutive terms is \(x\), which is constant, making it an arithmetic progression.


Final Answer: \[ \boxed{x, 2x, 3x, 4x, \dots} \]


% Quicktip
\begin{quicktipbox
In an A.P., the difference between consecutive terms is constant.
\end{quicktipbox Quick Tip: In an A.P., the difference between consecutive terms is constant.

Step 1: Identify the common difference in an A.P.

An arithmetic progression (A.P.) is a sequence of numbers where the difference between consecutive terms is constant.

Step 2: Analyze the given options.

Options (A), (B), and (C) are examples of sequences with constant differences between consecutive terms. However, in option (D), the terms \( 2^2, 4^2, 6^2, 8^2, \dots \) are squares of even numbers, and the difference between consecutive terms is not constant.


Final Answer: \[ \boxed{2^2, 4^2, 6^2, 8^2, \dots} \]


% Quicktip
\begin{quicktipbox
In an arithmetic progression (A.P.), the difference between consecutive terms is constant. The sequence of squares of numbers does not form an A.P.
\end{quicktipbox Quick Tip: In an arithmetic progression (A.P.), the difference between consecutive terms is constant. The sequence of squares of numbers does not form an A.P.

Step 1: Use the formula for the sum of an A.P.

The sum of the first \(n\) terms of an A.P. is given by: \[ S_n = \frac{n}{2} \cdot \left( 2a + (n-1) \cdot d \right) \]
where \(a\) is the first term, \(d\) is the common difference, and \(n\) is the number of terms.

Step 2: Substitute the given values.

We are given \(a = 1\), \(d = 3\), and \(n = 20\): \[ S_{20} = \frac{20}{2} \cdot \left( 2 \cdot 1 + (20-1) \cdot 3 \right) \]
Simplifying: \[ S_{20} = 10 \cdot \left( 2 + 57 \right) = 10 \cdot 59 = 590 \]


Final Answer: \[ \boxed{590} \]


% Quicktip
\begin{quicktipbox
The sum of the first \(n\) terms of an A.P. is calculated using \( S_n = \frac{n}{2} \cdot \left( 2a + (n-1) \cdot d \right) \).
\end{quicktipbox Quick Tip: The sum of the first \(n\) terms of an A.P. is calculated using \( S_n = \frac{n}{2} \cdot \left( 2a + (n-1) \cdot d \right) \).

Step 1: Use the Pythagorean identity.

The Pythagorean identity states that: \[ \sin^2 \theta + \cos^2 \theta = 1 \]
So, for \( \theta = 60^\circ \): \[ \sin^2 60^\circ + \cos^2 60^\circ = 1 \]

Step 2: Verify the other options.

- Option (B): \( \sin 90^\circ \cdot \cos 90^\circ = 1 \cdot 0 = 0 \), which is not equal to 1.
- Option (C): \( \sin^2 60^\circ = \left( \frac{\sqrt{3}}{2} \right)^2 = \frac{3}{4} \), which is not equal to 1.
- Option (D): \( \sin 45^\circ \cdot \frac{1}{\cos 45^\circ} = 1 \), but it is not written as 1 directly, while option (A) gives the exact result.


Final Answer: \[ \boxed{\sin^2 60^\circ + \cos^2 60^\circ = 1} \]


% Quicktip
\begin{quicktipbox
Remember the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \), which always holds true for any angle \( \theta \).
\end{quicktipbox Quick Tip: Remember the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \), which always holds true for any angle \( \theta \).

Step 1: Use the Pythagorean identity.

We know that the identity \( 1 + \tan^2 A = \sec^2 A \), so we can rewrite the expression as: \[ \cos^2 A(1 + \tan^2 A) = \cos^2 A \cdot \sec^2 A \]

Step 2: Simplify the expression.

Using \( \sec^2 A = \frac{1}{\cos^2 A} \), we get: \[ \cos^2 A \cdot \sec^2 A = \cos^2 A \cdot \frac{1}{\cos^2 A} = 1 \]


Final Answer: \[ \boxed{1} \]


% Quicktip
\begin{quicktipbox
Use the identity \( 1 + \tan^2 A = \sec^2 A \) to simplify trigonometric expressions.
\end{quicktipbox Quick Tip: Use the identity \( 1 + \tan^2 A = \sec^2 A \) to simplify trigonometric expressions.

Step 1: Use the standard value of \( \tan 30^\circ \).

From standard trigonometric tables or unit circle values, we know that: \[ \tan 30^\circ = \frac{1}{\sqrt{3}} \]


Final Answer: \[ \boxed{\frac{1}{\sqrt{3}}} \]


% Quicktip
\begin{quicktipbox
The value of \( \tan 30^\circ \) is \( \frac{1}{\sqrt{3}} \), which is a standard trigonometric value.
\end{quicktipbox Quick Tip: The value of \( \tan 30^\circ \) is \( \frac{1}{\sqrt{3}} \), which is a standard trigonometric value.

Step 1: Use the standard value of \( \cos 60^\circ \).

From standard trigonometric values, we know that: \[ \cos 60^\circ = \frac{1}{2} \]


Final Answer: \[ \boxed{\frac{1}{2}} \]


% Quicktip
\begin{quicktipbox
The value of \( \cos 60^\circ \) is \( \frac{1}{2} \), which is a standard trigonometric value.
\end{quicktipbox Quick Tip: The value of \( \cos 60^\circ \) is \( \frac{1}{2} \), which is a standard trigonometric value.

Step 1: Use standard values.

We know that: \[ \sin 90^\circ = 1 \quad and \quad \tan 45^\circ = 1 \]
So: \[ \sin^2 90^\circ = 1^2 = 1 \quad and \quad \tan^2 45^\circ = 1^2 = 1 \]

Step 2: Simplify the expression.

Now, substitute these values into the expression: \[ \sin^2 90^\circ - \tan^2 45^\circ = 1 - 1 = 0 \]


Final Answer: \[ \boxed{0} \]


% Quicktip
\begin{quicktipbox
Use standard trigonometric values like \( \sin 90^\circ = 1 \) and \( \tan 45^\circ = 1 \) to simplify expressions.
\end{quicktipbox Quick Tip: Use standard trigonometric values like \( \sin 90^\circ = 1 \) and \( \tan 45^\circ = 1 \) to simplify expressions.

Step 1: Use the distance formula.

The distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by the formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Step 2: Substitute the coordinates of the given points.

We are given the points \( (8 \sin 60^\circ, 0) \) and \( (0, 8 \cos 60^\circ) \). The coordinates of the first point are \( (8 \sin 60^\circ, 0) \), and the coordinates of the second point are \( (0, 8 \cos 60^\circ) \). Substituting into the distance formula: \[ d = \sqrt{(0 - 8 \sin 60^\circ)^2 + (8 \cos 60^\circ - 0)^2} \]

Step 3: Simplify the expression.

We know that \( \sin 60^\circ = \frac{\sqrt{3}}{2} \) and \( \cos 60^\circ = \frac{1}{2} \), so: \[ d = \sqrt{\left( 8 \times \frac{\sqrt{3}}{2} \right)^2 + \left( 8 \times \frac{1}{2} \right)^2} \] \[ d = \sqrt{\left( 4\sqrt{3} \right)^2 + (4)^2} = \sqrt{48 + 16} = \sqrt{64} = 8 \]


Final Answer: \[ \boxed{8} \]


% Quicktip
\begin{quicktipbox
Use the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \) to find the distance between two points in a coordinate plane.
\end{quicktipbox Quick Tip: Use the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \) to find the distance between two points in a coordinate plane.

Step 1: Use the distance formula from the origin.

The distance \( OP \) from the origin \( O(0, 0) \) to a point \( P(x, y) \) is given by the distance formula: \[ d = \sqrt{x^2 + y^2} \]

Step 2: Explanation.

This is derived from the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \), where the coordinates of \( O \) are \( (0, 0) \).


Final Answer: \[ \boxed{\sqrt{x^2 + y^2}} \]


% Quicktip
\begin{quicktipbox
The distance from the origin to any point \( (x, y) \) is given by \( \sqrt{x^2 + y^2} \), derived from the distance formula.
\end{quicktipbox Quick Tip: The distance from the origin to any point \( (x, y) \) is given by \( \sqrt{x^2 + y^2} \), derived from the distance formula.

Step 1: Understand the distance from the \( y \)-axis.

The distance of a point \( (x, y) \) from the \( y \)-axis is simply the absolute value of the \( x \)-coordinate.

Step 2: Apply the formula.

For the point \( (12, 14) \), the distance from the \( y \)-axis is: \[ Distance = |x| = |12| = 12 \]


Final Answer: \[ \boxed{12} \]


% Quicktip
\begin{quicktipbox
The distance of any point from the \( y \)-axis is the absolute value of the \( x \)-coordinate of the point.
\end{quicktipbox Quick Tip: The distance of any point from the \( y \)-axis is the absolute value of the \( x \)-coordinate of the point.

The ordinate of a point is the \(y\)-coordinate of that point.
For the point \( (-6, -8) \), the ordinate (or \(y\)-coordinate) is \( -8 \).


Final Answer: \[ \boxed{-8} \]


% Quicktip
\begin{quicktipbox
The ordinate of a point refers to the \(y\)-coordinate.
\end{quicktipbox Quick Tip: The ordinate of a point refers to the \(y\)-coordinate.

Step 1: Identify the coordinates.

The point \( (3, -4) \) has a positive \(x\)-coordinate and a negative \(y\)-coordinate.

Step 2: Analyze the quadrant.

In the Cartesian coordinate system:
- The first quadrant contains points where both \(x\) and \(y\) are positive.
- The second quadrant contains points where \(x\) is negative and \(y\) is positive.
- The third quadrant contains points where both \(x\) and \(y\) are negative.
- The fourth quadrant contains points where \(x\) is positive and \(y\) is negative.

Since the point \( (3, -4) \) has a positive \(x\)-coordinate and a negative \(y\)-coordinate, it lies in the fourth quadrant.


Final Answer: \[ \boxed{Fourth} \]


% Quicktip
\begin{quicktipbox
A point with a positive \(x\)-coordinate and a negative \(y\)-coordinate lies in the fourth quadrant.
\end{quicktipbox Quick Tip: A point with a positive \(x\)-coordinate and a negative \(y\)-coordinate lies in the fourth quadrant.

Step 1: Identify the coordinates for each option.

- The second quadrant contains points where \(x\) is negative and \(y\) is positive.

Step 2: Analyze the options.

- Option (A) \( (3, 2) \) lies in the first quadrant because \(x > 0\) and \(y > 0\).
- Option (B) \( (-3, 2) \) lies in the second quadrant because \(x < 0\) and \(y > 0\).
- Option (C) \( (3, -2) \) lies in the fourth quadrant because \(x > 0\) and \(y < 0\).
- Option (D) \( (-3, -2) \) lies in the third quadrant because \(x < 0\) and \(y < 0\).


Final Answer: \[ \boxed{(-3, 2)} \]


% Quicktip
\begin{quicktipbox
In the second quadrant, \(x\) is negative and \(y\) is positive.
\end{quicktipbox Quick Tip: In the second quadrant, \(x\) is negative and \(y\) is positive.

Step 1: Use the formula for the mid-point.

The formula for the mid-point of a line segment with endpoints \( (x_1, y_1) \) and \( (x_2, y_2) \) is: \[ \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]

Step 2: Substitute the given points.

We are given the points \( (4, -4) \) and \( (-4, 4) \). Applying the mid-point formula: \[ \left( \frac{4 + (-4)}{2}, \frac{-4 + 4}{2} \right) = \left( \frac{0}{2}, \frac{0}{2} \right) = (0, 0) \]


Final Answer: \[ \boxed{(0, 0)} \]


% Quicktip
\begin{quicktipbox
The mid-point of a line segment can be found using the formula \( \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \).
\end{quicktipbox Quick Tip: The mid-point of a line segment can be found using the formula \( \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \).

Step 1: Use the formula for the mid-point.

Let the coordinates of point \( B \) be \( (x_B, y_B) \). The formula for the mid-point of a line segment with endpoints \( A(x_A, y_A) \) and \( B(x_B, y_B) \) is: \[ \left( \frac{x_A + x_B}{2}, \frac{y_A + y_B}{2} \right) \]
We are given that the mid-point is \( (2, 4) \), and the coordinates of point \( A \) are \( (5, 7) \). Substituting into the formula: \[ \left( \frac{5 + x_B}{2}, \frac{7 + y_B}{2} \right) = (2, 4) \]

Step 2: Solve for \( x_B \) and \( y_B \).

For the \(x\)-coordinate: \[ \frac{5 + x_B}{2} = 2 \quad \Rightarrow \quad 5 + x_B = 4 \quad \Rightarrow \quad x_B = -1 \]

For the \(y\)-coordinate: \[ \frac{7 + y_B}{2} = 4 \quad \Rightarrow \quad 7 + y_B = 8 \quad \Rightarrow \quad y_B = 1 \]


Final Answer: \[ \boxed{(-1, 1)} \]


% Quicktip
\begin{quicktipbox
To find the missing point when the mid-point is known, use the mid-point formula and solve for the coordinates of the unknown point.
\end{quicktipbox Quick Tip: To find the missing point when the mid-point is known, use the mid-point formula and solve for the coordinates of the unknown point.

Step 1: Use the formula for the center of the circle.

The center of the circle lies at the mid-point of the diameter. The formula for the mid-point of a line segment joining the points \( (x_1, y_1) \) and \( (x_2, y_2) \) is: \[ \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]

Step 2: Substitute the given coordinates of the endpoints.

The endpoints of the diameter are \( (10, -6) \) and \( (-6, 10) \). Substituting these into the mid-point formula: \[ \left( \frac{10 + (-6)}{2}, \frac{-6 + 10}{2} \right) = \left( \frac{4}{2}, \frac{4}{2} \right) = (2, 2) \]


Final Answer: \[ \boxed{(2, 2)} \]


% Quicktip
\begin{quicktipbox
The center of a circle is the mid-point of the diameter. Use the mid-point formula to find the center.
\end{quicktipbox Quick Tip: The center of a circle is the mid-point of the diameter. Use the mid-point formula to find the center.

Step 1: Use the formula for the centroid.

The centroid of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) has coordinates: \[ \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \]

Step 2: Substitute the given points.

The given points are \( (4, 6) \), \( (0, 4) \), and \( (5, 5) \). Substituting these into the formula: \[ \left( \frac{4 + 0 + 5}{3}, \frac{6 + 4 + 5}{3} \right) = \left( \frac{9}{3}, \frac{15}{3} \right) = (3, 5) \]


Final Answer: \[ \boxed{(3, 5)} \]


% Quicktip
\begin{quicktipbox
The centroid of a triangle is found by averaging the \(x\)-coordinates and \(y\)-coordinates of the three vertices.
\end{quicktipbox Quick Tip: The centroid of a triangle is found by averaging the \(x\)-coordinates and \(y\)-coordinates of the three vertices.

Step 1: Identify fractions with terminating decimal expansions.

A fraction will have a terminating decimal expansion if and only if the denominator, after simplifying the fraction, has no prime factors other than 2 and 5.

Step 2: Simplify the given fractions.

- Option (A): \( \frac{14}{20 \times 32} = \frac{14}{640} = \frac{7}{320} \). The denominator \( 320 = 2^8 \times 5 \), which consists only of 2's and 5's, so this has a terminating decimal expansion.

- Option (B): \( \frac{9}{51 \times 72} = \frac{9}{3672} \). The denominator \( 3672 = 2^3 \times 3 \times 7 \), so it doesn't have only 2's and 5's. It does not have a terminating decimal.

- Option (C): \( \frac{8}{22 \times 32} = \frac{8}{704} = \frac{1}{88} \). The denominator \( 88 = 2^3 \times 11 \), which does not have only 2's and 5's, so this doesn't have a terminating decimal.

- Option (D): \( \frac{15}{22 \times 53} = \frac{15}{1166} \). The denominator \( 1166 = 2 \times 53 \times 11 \), which does not have only 2's and 5's, so this doesn't have a terminating decimal.


Final Answer: \[ \boxed{\frac{14}{20 \times 32}} \]


% Quicktip
\begin{quicktipbox
A fraction has a terminating decimal expansion if the simplified denominator contains only the prime factors 2 and 5.
\end{quicktipbox Quick Tip: A fraction has a terminating decimal expansion if the simplified denominator contains only the prime factors 2 and 5.

Step 1: Convert the decimal number to a fraction.

We are given \( 0.505 \), which can be written as \( \frac{505}{1000} \).

Step 2: Simplify the fraction.

Now, simplify \( \frac{505}{1000} \): \[ \frac{505}{1000} = \frac{101}{200} \]

Step 3: Express in the required form.

We need to express this fraction in the form \( \frac{p}{2^n \times 5^m} \). To do this, we factor the denominator 200: \[ 200 = 2^3 \times 5^2 \]
So, \( \frac{101}{200} = \frac{101}{2^3 \times 5^2} \), which matches option (A).


Final Answer: \[ \boxed{\frac{101}{2^1 \times 5^2}} \]


% Quicktip
\begin{quicktipbox
To convert decimals into fractions, multiply by powers of 10 and then simplify the resulting fraction.
\end{quicktipbox Quick Tip: To convert decimals into fractions, multiply by powers of 10 and then simplify the resulting fraction.

Step 1: Understand the division algorithm.

The division algorithm is given by: \[ a = bq + r \]
where \(a\) is the dividend, \(b\) is the divisor, \(q\) is the quotient, and \(r\) is the remainder.

Step 2: Substitute the known values.

We are given \( b = 4 \), \( q = 5 \), and \( r = 1 \). Substitute these values into the formula: \[ a = 4 \times 5 + 1 = 20 + 1 = 21 \]


Final Answer: \[ \boxed{21} \]


% Quicktip
\begin{quicktipbox
In the division algorithm, \( a = bq + r \), where \( a \) is the dividend, \( b \) is the divisor, \( q \) is the quotient, and \( r \) is the remainder.
\end{quicktipbox Quick Tip: In the division algorithm, \( a = bq + r \), where \( a \) is the dividend, \( b \) is the divisor, \( q \) is the quotient, and \( r \) is the remainder.

Step 1: Use the quadratic formula.

For a quadratic equation \( ax^2 + bx + c = 0 \), the roots (or zeroes) are given by the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For the equation \( 2x^2 - 4x - 6 = 0 \), we have \( a = 2 \), \( b = -4 \), and \( c = -6 \).

Step 2: Substitute the values into the formula.

Substitute \( a = 2 \), \( b = -4 \), and \( c = -6 \) into the quadratic formula: \[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \times 2 \times (-6)}}{2 \times 2} \] \[ x = \frac{4 \pm \sqrt{16 + 48}}{4} \] \[ x = \frac{4 \pm \sqrt{64}}{4} \] \[ x = \frac{4 \pm 8}{4} \]
Thus, the two roots are: \[ x = \frac{4 + 8}{4} = 3 \quad and \quad x = \frac{4 - 8}{4} = -1 \]


Final Answer: \[ \boxed{1, -3} \]


% Quicktip
\begin{quicktipbox
The roots of a quadratic equation can be found using the quadratic formula. Make sure to calculate the discriminant \( \Delta = b^2 - 4ac \) first.
\end{quicktipbox Quick Tip: The roots of a quadratic equation can be found using the quadratic formula. Make sure to calculate the discriminant \( \Delta = b^2 - 4ac \) first.

Step 1: Use the degree property of polynomials.

The degree of the product of two polynomials is the sum of their degrees.

Step 2: Determine the degree of each polynomial.

The degree of \( (x^3 + x^2 + 2x + 1) \) is 3, as the highest power of \(x\) is 3.
The degree of \( (x^2 + 2x + 1) \) is 2, as the highest power of \(x\) is 2.

Step 3: Add the degrees.

The degree of the product is \( 3 + 2 = 5 \).


Final Answer: \[ \boxed{5} \]


% Quicktip
\begin{quicktipbox
The degree of the product of two polynomials is the sum of their degrees.
\end{quicktipbox Quick Tip: The degree of the product of two polynomials is the sum of their degrees.

Step 1: Definition of a polynomial.

A polynomial is an expression that involves sums and differences of powers of a variable, with non-negative integer exponents.

Step 2: Identify which expression is not a polynomial.

- Option (A) \( x^2 - 7 \) is a polynomial because it involves a non-negative integer exponent.
- Option (B) \( 2x^2 + 7x + 6 \) is a polynomial because it involves non-negative integer exponents.
- Option (C) \( \frac{1}{2} x^2 + \frac{1}{2} x + 4 \) is a polynomial because it involves non-negative integer exponents.
- Option (D) \( \frac{4}{x} \) is not a polynomial because \( x \) is in the denominator, which means the exponent is negative, violating the definition of a polynomial.


Final Answer: \[ \boxed{\frac{4}{x}} \]


% Quicktip
\begin{quicktipbox
Polynomials cannot have negative exponents or variables in the denominator.
\end{quicktipbox Quick Tip: Polynomials cannot have negative exponents or variables in the denominator.

Step 1: Use the factorization of a quadratic equation.

A quadratic equation with roots \( p \) and \( q \) can be written as: \[ (x - p)(x - q) \]

Step 2: Form the quadratic equation.

The roots are \( 2 \) and \( -2 \). The corresponding quadratic equation is: \[ (x - 2)(x + 2) = x^2 - 4 \]

Step 3: Verify the other options.

- Option (A) \( x^2 + 4 \) does not have real roots because the discriminant is negative.
- Option (C) \( x^2 - 2x + 4 \) has complex roots because the discriminant is negative.
- Option (D) \( x^2 + \sqrt{5} \) does not have real roots because the discriminant is negative.


Final Answer: \[ \boxed{x^2 - 4} \]


% Quicktip
\begin{quicktipbox
For a quadratic equation with roots \( p \) and \( q \), the equation is \( (x - p)(x - q) \).
\end{quicktipbox Quick Tip: For a quadratic equation with roots \( p \) and \( q \), the equation is \( (x - p)(x - q) \).

The sum and product of the roots \( \alpha \) and \( \beta \) of a quadratic equation \( ax^2 + bx + c \) are given by Vieta's formulas:
- Sum of the roots \( \alpha + \beta = -\frac{b}{a} \)
- Product of the roots \( \alpha \beta = \frac{c}{a} \)

For the polynomial \( x^2 + 7x + 10 \), \( a = 1 \), \( b = 7 \), and \( c = 10 \). Therefore: \[ \alpha \beta = \frac{c}{a} = \frac{10}{1} = 10 \]


Final Answer: \[ \boxed{10} \]


% Quicktip
\begin{quicktipbox
The product of the roots of a quadratic equation \( ax^2 + bx + c \) is \( \frac{c}{a} \).
\end{quicktipbox Quick Tip: The product of the roots of a quadratic equation \( ax^2 + bx + c \) is \( \frac{c}{a} \).

We know the following standard trigonometric values:
- \( \sin 30^\circ = \frac{1}{2} \), \( \cos 30^\circ = \frac{\sqrt{3}}{2} \)
- \( \sin 60^\circ = \frac{\sqrt{3}}{2} \), \( \cos 60^\circ = \frac{1}{2} \)

Step 1: Substitute the values.

Substitute these values into the expression: \[ \left( \sin 30^\circ + \cos 30^\circ \right) - \left( \sin 60^\circ + \cos 60^\circ \right) = \left( \frac{1}{2} + \frac{\sqrt{3}}{2} \right) - \left( \frac{\sqrt{3}}{2} + \frac{1}{2} \right) \] \[ = \frac{1}{2} + \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} - \frac{1}{2} = 0 \]


Final Answer: \[ \boxed{0} \]


% Quicktip
\begin{quicktipbox
When simplifying trigonometric expressions, substitute known values and simplify terms.
\end{quicktipbox Quick Tip: When simplifying trigonometric expressions, substitute known values and simplify terms.

Step 1: Use the fact that -4 is a root.

Since \( -4 \) is a root of the quadratic equation, we substitute \( x = -4 \) into the quadratic equation \( (k - 1)x^2 + kx + 1 = 0 \).

Step 2: Substitute \( x = -4 \).

Substitute \( x = -4 \) into the equation: \[ (k - 1)(-4)^2 + k(-4) + 1 = 0 \] \[ (k - 1) \cdot 16 - 4k + 1 = 0 \] \[ 16k - 16 - 4k + 1 = 0 \] \[ 12k - 15 = 0 \] \[ 12k = 15 \] \[ k = \frac{15}{12} = \frac{5}{4} \]


Final Answer: \[ \boxed{\frac{5}{4}} \]


% Quicktip
\begin{quicktipbox
To find the value of \( k \) in a quadratic equation, substitute the known root into the equation and solve for \( k \).
\end{quicktipbox Quick Tip: To find the value of \( k \) in a quadratic equation, substitute the known root into the equation and solve for \( k \).

Step 1: Condition for real and equal roots.

The condition for a quadratic equation \( ax^2 + bx + c = 0 \) to have real and equal roots is that the discriminant \( \Delta \) must be zero. The discriminant is given by: \[ \Delta = b^2 - 4ac \]

Step 2: Apply this condition to the given equation.

For the equation \( kx^2 - 6x + 1 = 0 \), we have \( a = k \), \( b = -6 \), and \( c = 1 \). The discriminant \( \Delta \) becomes: \[ \Delta = (-6)^2 - 4(k)(1) = 36 - 4k \]

Step 3: Set the discriminant equal to zero.

For the roots to be real and equal, the discriminant must be zero: \[ 36 - 4k = 0 \] \[ 4k = 36 \] \[ k = \frac{36}{4} = 9 \]


Final Answer: \[ \boxed{9} \]


% Quicktip
\begin{quicktipbox
For real and equal roots, set the discriminant \( \Delta = 0 \) and solve for \( k \).
\end{quicktipbox Quick Tip: For real and equal roots, set the discriminant \( \Delta = 0 \) and solve for \( k \).

Step 1: Use the relationship between zeroes and factors.

If \( 2 \) is a zero of the polynomial \( p(x) \), then \( (x - 2) \) is a factor of \( p(x) \). This is because if \( x = 2 \) satisfies the equation \( p(x) = 0 \), then \( x - 2 \) must divide the polynomial.

Step 2: Identify the factor.

Given that one of the zeroes is 2, the factor must be \( x - 2 \).


Final Answer: \[ \boxed{x - 2} \]


% Quicktip
\begin{quicktipbox
If \( \alpha \) is a zero of a polynomial \( p(x) \), then \( (x - \alpha) \) is a factor of \( p(x) \).
\end{quicktipbox Quick Tip: If \( \alpha \) is a zero of a polynomial \( p(x) \), then \( (x - \alpha) \) is a factor of \( p(x) \).

The sum and product of the roots \( \alpha \) and \( \beta \) of the quadratic equation \( ax^2 + bx + c \) are given by Vieta's formulas:
- The sum of the roots \( \alpha + \beta = -\frac{b}{a} \)
- The product of the roots \( \alpha \beta = \frac{c}{a} \)

For the polynomial \( x^2 + ax + b \), the coefficient of \( x^2 \) is \( 1 \), \( a \) is the coefficient of \( x \), and \( b \) is the constant term. Therefore, the product of the roots is: \[ \alpha \beta = \frac{b}{1} = b \]


Final Answer: \[ \boxed{b} \]


% Quicktip
\begin{quicktipbox
For a quadratic equation \( ax^2 + bx + c = 0 \), the product of the roots is \( \frac{c}{a} \).
\end{quicktipbox Quick Tip: For a quadratic equation \( ax^2 + bx + c = 0 \), the product of the roots is \( \frac{c}{a} \).

A quadratic equation is one where the highest power of \(x\) is 2. Let's analyze each option:

- Option (A) \( (x + 3)(x - 3) = x^2 - 4x^3 \): This is not a quadratic equation because the degree of the right side is 3.

- Option (B) \( (x + 3)^3 = 4(x + 4) \): This is not a quadratic equation because the degree of the left side is 3.

- Option (C) \( (2x - 2)^2 = 4x^2 + 7 \): Expanding the left side: \[ (2x - 2)^2 = 4x^2 - 8x + 4 \]
This is a quadratic equation because the highest power of \(x\) is 2.

- Option (D) \( 4x + \frac{1}{4}x = 4x \): This is a linear equation, not a quadratic equation.


Final Answer: \[ \boxed{(2x - 2)^2 = 4x^2 + 7} \]


% Quicktip
\begin{quicktipbox
A quadratic equation has the highest power of \(x\) as 2.
\end{quicktipbox Quick Tip: A quadratic equation has the highest power of \(x\) as 2.

A quadratic equation has the highest power of \(x\) as 2. Let's analyze each option:

- Option (A) \( 5x^2 - x^2 + 3 = 4x^2 + 3 \), which is a quadratic equation since the highest power of \(x\) is 2.

- Option (B) \( x^3 - x^2 = (x - 1)^3 \): This equation has \(x^3\) on the left-hand side, meaning the degree of the equation is 3, so it is not a quadratic equation.

- Option (C) \( (x + 3)^2 = 3(x^2 - 5) \) is a quadratic equation after simplification.

- Option (D) \( (5x^3 - x^2 + 3) \) is not a quadratic equation because of the \(x^3\) term.


Final Answer: \[ \boxed{x^3 - x^2 = (x - 1)^3} \]


% Quicktip
\begin{quicktipbox
A quadratic equation has the highest power of \(x\) as 2. Equations with higher powers (like \(x^3\)) are not quadratic equations.
\end{quicktipbox Quick Tip: A quadratic equation has the highest power of \(x\) as 2. Equations with higher powers (like \(x^3\)) are not quadratic equations.

Step 1: Formula for the discriminant.

The discriminant \( \Delta \) of a quadratic equation \( ax^2 + bx + c = 0 \) is given by the formula: \[ \Delta = b^2 - 4ac \]

Step 2: Identify the values of \(a\), \(b\), and \(c\).

For the equation \( 2x^2 - 7x + 6 = 0 \), we have:
- \( a = 2 \)
- \( b = -7 \)
- \( c = 6 \)

Step 3: Substitute the values into the discriminant formula.

Now substitute \( a = 2 \), \( b = -7 \), and \( c = 6 \) into the formula for the discriminant: \[ \Delta = (-7)^2 - 4(2)(6) \] \[ \Delta = 49 - 48 \] \[ \Delta = 1 \]


Final Answer: \[ \boxed{1} \]


% Quicktip
\begin{quicktipbox
The discriminant \( \Delta \) of a quadratic equation \( ax^2 + bx + c = 0 \) is calculated using the formula \( \Delta = b^2 - 4ac \).
\end{quicktipbox Quick Tip: The discriminant \( \Delta \) of a quadratic equation \( ax^2 + bx + c = 0 \) is calculated using the formula \( \Delta = b^2 - 4ac \).

Step 1: Understand the equation.

The equation \( x - 2 = 0 \) represents a vertical line where the value of \( x \) is always equal to 2. This means that all points on the graph of this equation have the \( x \)-coordinate as 2.

Step 2: Check the given points.

- Option (A) \( (2, 0) \): This satisfies the equation because \( x = 2 \).
- Option (B) \( (2, 1) \): This also satisfies the equation because \( x = 2 \).
- Option (C) \( (2, 2) \): This satisfies the equation because \( x = 2 \).
- Option (D) all of these: Since all the points satisfy the equation, this is the correct answer.


Final Answer: \[ \boxed{all of these} \]


% Quicktip
\begin{quicktipbox
The equation \( x - 2 = 0 \) represents a vertical line where \( x = 2 \) for all points on the line.
\end{quicktipbox Quick Tip: The equation \( x - 2 = 0 \) represents a vertical line where \( x = 2 \) for all points on the line.

Step 1: Use the property of Arithmetic Progression (A.P.).

In an A.P., the middle term is the average of the first and third terms. Therefore, the following condition holds: \[ 2P + 1 = \frac{(P + 1) + (4P - 1)}{2} \]

Step 2: Simplify the equation.
\[ 2P + 1 = \frac{(P + 1) + (4P - 1)}{2} \] \[ 2P + 1 = \frac{5P}{2} \]
Multiply both sides by 2: \[ 4P + 2 = 5P \] \[ 4P - 5P = -2 \] \[ -P = -2 \] \[ P = 2 \]


Final Answer: \[ \boxed{3} \]


% Quicktip
\begin{quicktipbox
In an A.P., the middle term is the average of the first and third terms. Use this property to solve for the unknown term.
\end{quicktipbox Quick Tip: In an A.P., the middle term is the average of the first and third terms. Use this property to solve for the unknown term.

The common difference \( d \) in an arithmetic progression (A.P.) is the difference between any two consecutive terms.

Step 1: Find the common difference.

For the A.P. \( 1, 5, 9, \dots \), the common difference \( d \) is: \[ d = 5 - 1 = 4 \]
or \[ d = 9 - 5 = 4 \]

Step 2: Conclusion.

The common difference of this A.P. is 4.


Final Answer: \[ \boxed{4} \]


% Quicktip
\begin{quicktipbox
The common difference \( d \) in an A.P. is found by subtracting any term from the next term in the sequence.
\end{quicktipbox Quick Tip: The common difference \( d \) in an A.P. is found by subtracting any term from the next term in the sequence.

The \( n \)-th term of an arithmetic progression is given by: \[ T_n = a + (n-1)d \]
where \( a \) is the first term, \( d \) is the common difference, and \( n \) is the term number.

Step 1: Identify the values.

For the A.P. \( 5, 8, 11, 14, \dots \):
- First term \( a = 5 \)
- Common difference \( d = 3 \)
- We are asked to find the term where \( T_n = 38 \).

Step 2: Set up the equation.

Substitute \( T_n = 38 \), \( a = 5 \), and \( d = 3 \) into the formula: \[ 38 = 5 + (n - 1) \cdot 3 \]

Step 3: Solve for \( n \).

Simplify the equation: \[ 38 = 5 + 3(n - 1) \] \[ 38 - 5 = 3(n - 1) \] \[ 33 = 3(n - 1) \] \[ \frac{33}{3} = n - 1 \] \[ 11 = n - 1 \] \[ n = 12 \]


Final Answer: \[ \boxed{12th} \]


% Quicktip
\begin{quicktipbox
Use the formula for the \( n \)-th term of an A.P. to find the term number by substituting the given value and solving for \( n \).
\end{quicktipbox Quick Tip: Use the formula for the \( n \)-th term of an A.P. to find the term number by substituting the given value and solving for \( n \).

This is a standard trigonometric identity. The identity is: \[ \sin(90^\circ - A) = \cos A \]

Step 1: State the identity.

This identity is known as the co-function identity for sine and cosine.

Step 2: Conclusion.

Therefore, \( \sin(90^\circ - A) = \cos A \).


Final Answer: \[ \boxed{\cos A} \]


% Quicktip
\begin{quicktipbox
The co-function identity states that \( \sin(90^\circ - A) = \cos A \).
\end{quicktipbox Quick Tip: The co-function identity states that \( \sin(90^\circ - A) = \cos A \).

We are given \( \alpha = \beta = 60^\circ \), so we need to calculate \( \cos(\alpha - \beta) \).

Step 1: Use the cosine of difference identity.

The formula for the cosine of the difference of two angles is: \[ \cos(A - B) = \cos A \cdot \cos B + \sin A \cdot \sin B \]

Step 2: Substitute \( \alpha = \beta = 60^\circ \).

Substitute \( \alpha = \beta = 60^\circ \) into the formula: \[ \cos(60^\circ - 60^\circ) = \cos 60^\circ \cdot \cos 60^\circ + \sin 60^\circ \cdot \sin 60^\circ \]

We know that: \[ \cos 60^\circ = \frac{1}{2}, \quad \sin 60^\circ = \frac{\sqrt{3}}{2} \]

Step 3: Calculate the value.
\[ \cos(0^\circ) = \left( \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2 = \frac{1}{4} + \frac{3}{4} = 1 \]


Final Answer: \[ \boxed{1} \]


% Quicktip
\begin{quicktipbox
For \( \alpha = \beta \), \( \cos(\alpha - \beta) = \cos 0^\circ = 1 \).
\end{quicktipbox Quick Tip: For \( \alpha = \beta \), \( \cos(\alpha - \beta) = \cos 0^\circ = 1 \).

We are given \( \theta = 45^\circ \). Using standard trigonometric values:
\[ \sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}} \]

Step 1: Calculate \( \sin \theta + \cos \theta \).
\[ \sin 45^\circ + \cos 45^\circ = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \]


Final Answer: \[ \boxed{\sqrt{2}} \]


% Quicktip
\begin{quicktipbox
For \( \theta = 45^\circ \), \( \sin \theta + \cos \theta = \sqrt{2} \).
\end{quicktipbox Quick Tip: For \( \theta = 45^\circ \), \( \sin \theta + \cos \theta = \sqrt{2} \).

We are given \( A = 30^\circ \). The expression \( \frac{2 \tan A}{1 - \tan^2 A} \) is the formula for \( \tan 2A \).

Step 1: Use the double angle formula for tangent.

The double angle formula for tangent is: \[ \tan 2A = \frac{2 \tan A}{1 - \tan^2 A} \]
For \( A = 30^\circ \), we have: \[ \tan 60^\circ = \frac{2 \tan 30^\circ}{1 - \tan^2 30^\circ} \]

Step 2: Apply the known value of \( \tan 30^\circ \).

We know that \( \tan 30^\circ = \frac{1}{\sqrt{3}} \), and using the formula, we get: \[ \tan 60^\circ = 2 \tan 30^\circ \]


Final Answer: \[ \boxed{2 \tan 60^\circ} \]


% Quicktip
\begin{quicktipbox
The formula \( \frac{2 \tan A}{1 - \tan^2 A} \) is used to find \( \tan 2A \).
\end{quicktipbox Quick Tip: The formula \( \frac{2 \tan A}{1 - \tan^2 A} \) is used to find \( \tan 2A \).

We are given \( \tan \theta = \frac{12}{5} \). To find \( \sin \theta \), we can use the identity: \[ \tan \theta = \frac{\sin \theta}{\cos \theta} \]
We can also use the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \).

Step 1: Use the identity for tangent.

We are given \( \tan \theta = \frac{12}{5} \), so: \[ \frac{\sin \theta}{\cos \theta} = \frac{12}{5} \]
This implies: \[ \sin \theta = \frac{12}{5} \cos \theta \]

Step 2: Use the Pythagorean identity.

Substitute \( \sin \theta = \frac{12}{5} \cos \theta \) into the identity \( \sin^2 \theta + \cos^2 \theta = 1 \): \[ \left( \frac{12}{5} \cos \theta \right)^2 + \cos^2 \theta = 1 \] \[ \frac{144}{25} \cos^2 \theta + \cos^2 \theta = 1 \] \[ \left( \frac{144}{25} + 1 \right) \cos^2 \theta = 1 \] \[ \frac{169}{25} \cos^2 \theta = 1 \] \[ \cos^2 \theta = \frac{25}{169} \] \[ \cos \theta = \frac{5}{13} \]

Step 3: Find \( \sin \theta \).

Now that we know \( \cos \theta = \frac{5}{13} \), we can find \( \sin \theta \) using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \): \[ \sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{25}{169} = \frac{144}{169} \] \[ \sin \theta = \frac{12}{13} \]


Final Answer: \[ \boxed{\frac{12}{13}} \]


% Quicktip
\begin{quicktipbox
Use the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \) to solve for \( \sin \theta \) and \( \cos \theta \).
\end{quicktipbox Quick Tip: Use the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \) to solve for \( \sin \theta \) and \( \cos \theta \).

We are given the expression: \[ \frac{\cos 59^\circ \cdot \tan 80^\circ}{\sin 31^\circ \cdot \cot 10^\circ} \]

Step 1: Use trigonometric identities.

First, simplify the expression using known values or identities:
- \( \tan 80^\circ = \cot 10^\circ \)
- \( \sin 31^\circ = \cos 59^\circ \) (because \( 31^\circ + 59^\circ = 90^\circ \))

Thus, the expression becomes: \[ \frac{\cos 59^\circ \cdot \cot 10^\circ}{\cos 59^\circ \cdot \cot 10^\circ} \]

Step 2: Simplify.

The terms cancel out, so we are left with: \[ 1 \]


Final Answer: \[ \boxed{1} \]


% Quicktip
\begin{quicktipbox
When simplifying trigonometric expressions, look for pairs of functions that cancel out, such as \( \tan \theta \) and \( \cot (90^\circ - \theta) \).
\end{quicktipbox Quick Tip: When simplifying trigonometric expressions, look for pairs of functions that cancel out, such as \( \tan \theta \) and \( \cot (90^\circ - \theta) \).

We are given that \( \tan 25^\circ \times \tan 65^\circ = \sin A \).

Step 1: Use a known trigonometric identity.

From the identity \( \tan x \times \tan (90^\circ - x) = 1 \), we know that: \[ \tan 25^\circ \times \tan 65^\circ = 1 \]

Thus, we have: \[ \sin A = 1 \]

Step 2: Solve for \( A \).

The sine of 90° is 1, so: \[ A = 90^\circ \]


Final Answer: \[ \boxed{90^\circ} \]


% Quicktip
\begin{quicktipbox
Use the identity \( \tan x \times \tan (90^\circ - x) = 1 \) to simplify expressions involving tangent functions.
\end{quicktipbox Quick Tip: Use the identity \( \tan x \times \tan (90^\circ - x) = 1 \) to simplify expressions involving tangent functions.

We are given \( \cos \theta = x \), and we are asked to find \( \tan \theta \).

Step 1: Use the trigonometric identity for tangent.

The identity for tangent is: \[ \tan \theta = \frac{\sin \theta}{\cos \theta} \]

Step 2: Use the Pythagorean identity.

We know the Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \]

Substitute \( \cos \theta = x \) into the identity: \[ \sin^2 \theta = 1 - x^2 \] \[ \sin \theta = \sqrt{1 - x^2} \]

Step 3: Find \( \tan \theta \).

Now, substitute \( \sin \theta = \sqrt{1 - x^2} \) and \( \cos \theta = x \) into the formula for \( \tan \theta \): \[ \tan \theta = \frac{\sqrt{1 - x^2}}{x} \]


Final Answer: \[ \boxed{\frac{x}{\sqrt{1 - x^2}}} \]


% Quicktip
\begin{quicktipbox
Use the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \) to express \( \sin \theta \) in terms of \( \cos \theta \), and then use the identity for \( \tan \theta \).
\end{quicktipbox Quick Tip: Use the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \) to express \( \sin \theta \) in terms of \( \cos \theta \), and then use the identity for \( \tan \theta \).

We are given the expression \( (1 - \cos^4 \theta) \) and need to simplify it.

Step 1: Factor the expression.

The given expression \( 1 - \cos^4 \theta \) is a difference of squares, so we can factor it as: \[ 1 - \cos^4 \theta = (1 - \cos^2 \theta)(1 + \cos^2 \theta) \]

Step 2: Express the result.

The expression \( 1 - \cos^2 \theta \) is equivalent to \( \sin^2 \theta \), so: \[ 1 - \cos^4 \theta = \sin^2 \theta (1 + \cos^2 \theta) \]


Final Answer: \[ \boxed{\sin^2 \theta (1 + \cos^2 \theta)} \]


% Quicktip
\begin{quicktipbox
Use the difference of squares formula to factor expressions like \( 1 - \cos^4 \theta \).
\end{quicktipbox Quick Tip: Use the difference of squares formula to factor expressions like \( 1 - \cos^4 \theta \).

A point on the \( y \)-axis will have \( x = 0 \) (since the \( x \)-coordinate of every point on the \( y \)-axis is 0). Thus, the form of a point on the \( y \)-axis is \( (0, y) \).


Final Answer: \[ \boxed{(0, y)} \]


% Quicktip
\begin{quicktipbox
A point on the \( y \)-axis has \( x = 0 \) and its coordinates are of the form \( (0, y) \).
\end{quicktipbox Quick Tip: A point on the \( y \)-axis has \( x = 0 \) and its coordinates are of the form \( (0, y) \).

We are given that the quadratic polynomial has zeros 3 and -10. The general form of a quadratic polynomial with roots \( r_1 \) and \( r_2 \) is given by: \[ P(x) = (x - r_1)(x - r_2) \]

Step 1: Write the polynomial with the given zeros.

Using the zeros 3 and -10, the polynomial is: \[ P(x) = (x - 3)(x + 10) \]

Step 2: Expand the product.
\[ P(x) = x^2 + 10x - 3x - 30 \] \[ P(x) = x^2 + 7x - 30 \]

Step 3: Conclusion.

Thus, the correct polynomial is \( x^2 + 7x - 30 \), which corresponds to option (B).


Final Answer: \[ \boxed{x^2 + 7x - 30} \]


% Quicktip
\begin{quicktipbox
To form a quadratic polynomial with given zeros, use the fact that the polynomial is the product of \( (x - r_1)(x - r_2) \), where \( r_1 \) and \( r_2 \) are the zeros.
\end{quicktipbox Quick Tip: To form a quadratic polynomial with given zeros, use the fact that the polynomial is the product of \( (x - r_1)(x - r_2) \), where \( r_1 \) and \( r_2 \) are the zeros.

We are given that the sum of the zeros \( \alpha + \beta = 3 \) and their product \( \alpha \beta = -2 \).

The general form of a quadratic polynomial with zeros \( \alpha \) and \( \beta \) is: \[ P(x) = x^2 - (\alpha + \beta)x + \alpha \beta \]

Step 1: Substitute the sum and product of the zeros.

Substituting \( \alpha + \beta = 3 \) and \( \alpha \beta = -2 \) into the polynomial: \[ P(x) = x^2 - 3x - 2 \]

Step 2: Conclusion.

Thus, the correct quadratic polynomial is \( x^2 - 3x - 2 \), which corresponds to option (A).


Final Answer: \[ \boxed{x^2 - 3x - 2} \]


% Quicktip
\begin{quicktipbox
For a quadratic polynomial with given sum and product of zeros, use the formula \( P(x) = x^2 - (\alpha + \beta)x + \alpha \beta \).
\end{quicktipbox Quick Tip: For a quadratic polynomial with given sum and product of zeros, use the formula \( P(x) = x^2 - (\alpha + \beta)x + \alpha \beta \).

The degree of the quotient when one polynomial is divided by another is the difference between the degrees of the two polynomials.

Step 1: Find the degree of \( p(x) \).

The degree of \( p(x) = x^4 - 2x^3 + 17x^2 - 4x + 30 \) is 4 because the highest power of \( x \) is \( x^4 \).

Step 2: Find the degree of \( q(x) \).

The degree of \( q(x) = x + 2 \) is 1 because the highest power of \( x \) is \( x^1 \).

Step 3: Calculate the degree of the quotient.

The degree of the quotient is the difference between the degrees of \( p(x) \) and \( q(x) \): \[ Degree of quotient = 4 - 1 = 3 \]

Thus, the degree of the quotient is 3.


Final Answer: \[ \boxed{3} \]


% Quicktip
\begin{quicktipbox
The degree of the quotient of two polynomials is the difference between their degrees.
\end{quicktipbox Quick Tip: The degree of the quotient of two polynomials is the difference between their degrees.

We are given the system of equations: \[ x + 2y + 3 = 0 \quad (Equation 1) \] \[ 3x + 6y + 9 = 0 \quad (Equation 2) \]

Step 1: Simplify the equations.

Notice that Equation 2 is a multiple of Equation 1. Specifically, Equation 2 is obtained by multiplying Equation 1 by 3: \[ 3(x + 2y + 3) = 3x + 6y + 9 = 0 \]
Thus, both equations are essentially the same.

Step 2: Conclusion.

Since both equations represent the same line, there are infinitely many solutions.


Final Answer: \[ \boxed{Infinitely many solutions} \]


% Quicktip
\begin{quicktipbox
If two equations in a system are multiples of each other, the system has infinitely many solutions (the equations represent the same line).
\end{quicktipbox Quick Tip: If two equations in a system are multiples of each other, the system has infinitely many solutions (the equations represent the same line).

When two linear equations are parallel, their graphs do not intersect. This means that there is no point where both equations are satisfied simultaneously, meaning there is no solution.

Step 1: Understand the nature of parallel lines.

Parallel lines have the same slope but different y-intercepts. Since they never meet, they do not have any common points, and hence, no solution.


Final Answer: \[ \boxed{none of these} \]


% Quicktip
\begin{quicktipbox
When two linear equations are parallel, the system has no solutions because their graphs do not intersect.
\end{quicktipbox Quick Tip: When two linear equations are parallel, the system has no solutions because their graphs do not intersect.

We are given the system of linear equations: \[ 5x - 4y + 8 = 0 \quad (Equation 1) \] \[ 7x + 6y - 9 = 0 \quad (Equation 2) \]

To determine the nature of the system (consistent or inconsistent), we can use the method of comparing slopes. For a system to be consistent, the lines must intersect at exactly one point, and for the system to be inconsistent, the lines must be parallel.

Step 1: Write the equations in slope-intercept form.

Solve Equation 1 for \( y \): \[ 5x - 4y + 8 = 0 \quad \Rightarrow \quad -4y = -5x - 8 \quad \Rightarrow \quad y = \frac{5}{4}x + 2 \]

Solve Equation 2 for \( y \): \[ 7x + 6y - 9 = 0 \quad \Rightarrow \quad 6y = -7x + 9 \quad \Rightarrow \quad y = \frac{-7}{6}x + \frac{3}{2} \]

Step 2: Compare the slopes.

- The slope of the first equation is \( \frac{5}{4} \).
- The slope of the second equation is \( \frac{-7}{6} \).

Since the slopes are not equal, the lines are not parallel, and therefore, the system has no solution.


Final Answer: \[ \boxed{inconsistent} \]


% Quicktip
\begin{quicktipbox
If the slopes of two linear equations are different, the system is inconsistent and has no solution.
\end{quicktipbox Quick Tip: If the slopes of two linear equations are different, the system is inconsistent and has no solution.

We are given the quadratic equation \( 3x^2 - 5x + 2 = 0 \) with roots \( \alpha \) and \( \beta \).

Step 1: Use Vieta's formulas.

From Vieta’s formulas, for the equation \( ax^2 + bx + c = 0 \), the sum and product of the roots \( \alpha \) and \( \beta \) are: \[ \alpha + \beta = -\frac{b}{a}, \quad \alpha \beta = \frac{c}{a} \]

For the given equation \( 3x^2 - 5x + 2 = 0 \):
- \( a = 3 \), \( b = -5 \), \( c = 2 \)
Thus: \[ \alpha + \beta = -\frac{-5}{3} = \frac{5}{3}, \quad \alpha \beta = \frac{2}{3} \]

Step 2: Use the identity for \( \alpha^2 + \beta^2 \).

We can express \( \alpha^2 + \beta^2 \) in terms of \( \alpha + \beta \) and \( \alpha \beta \) as: \[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta \]
Substitute the values for \( \alpha + \beta \) and \( \alpha \beta \): \[ \alpha^2 + \beta^2 = \left( \frac{5}{3} \right)^2 - 2 \times \frac{2}{3} \] \[ \alpha^2 + \beta^2 = \frac{25}{9} - \frac{4}{3} \]
Now, convert \( \frac{4}{3} \) to have a denominator of 9: \[ \alpha^2 + \beta^2 = \frac{25}{9} - \frac{12}{9} = \frac{13}{9} \]


Final Answer: \[ \boxed{\frac{13}{9}} \]


% Quicktip
\begin{quicktipbox
To find \( \alpha^2 + \beta^2 \), use the identity \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta \) and apply Vieta's formulas.
\end{quicktipbox Quick Tip: To find \( \alpha^2 + \beta^2 \), use the identity \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta \) and apply Vieta's formulas.

We are given that one root of the quadratic equation \( 2x^2 - 7x - p = 0 \) is 2. We need to find the value of \( p \).

Step 1: Use the given root in the equation.

Substitute \( x = 2 \) into the quadratic equation: \[ 2(2)^2 - 7(2) - p = 0 \] \[ 2 \times 4 - 14 - p = 0 \] \[ 8 - 14 - p = 0 \] \[ -6 - p = 0 \]

Step 2: Solve for \( p \).

Solving for \( p \): \[ p = -6 \]


Final Answer: \[ \boxed{-6} \]


% Quicktip
\begin{quicktipbox
To find the unknown constant in a quadratic equation, substitute one of the known roots into the equation and solve for the unknown.
\end{quicktipbox Quick Tip: To find the unknown constant in a quadratic equation, substitute one of the known roots into the equation and solve for the unknown.

We are given the quadratic equation \( 2x^2 - x - 6 = 0 \) and one of its roots is \( -\frac{3}{2} \). We need to find the other root.

Step 1: Use Vieta's formulas.

From Vieta’s formulas, for a quadratic equation \( ax^2 + bx + c = 0 \), the sum and product of the roots \( r_1 \) and \( r_2 \) are: \[ r_1 + r_2 = -\frac{b}{a}, \quad r_1 \cdot r_2 = \frac{c}{a} \]

For the equation \( 2x^2 - x - 6 = 0 \), \( a = 2 \), \( b = -1 \), and \( c = -6 \).

Step 2: Find the sum and product of the roots.

Using Vieta's formulas: \[ r_1 + r_2 = -\frac{-1}{2} = \frac{1}{2}, \quad r_1 \cdot r_2 = \frac{-6}{2} = -3 \]

Step 3: Use the known root.

We know one root \( r_1 = -\frac{3}{2} \). Now, substitute this into the sum and product formulas: \[ r_1 + r_2 = \frac{1}{2} \quad \Rightarrow \quad -\frac{3}{2} + r_2 = \frac{1}{2} \] \[ r_2 = \frac{1}{2} + \frac{3}{2} = 2 \]

Step 4: Conclusion.

Thus, the other root is \( r_2 = \frac{3}{2} \).


Final Answer: \[ \boxed{\frac{3}{2}} \]


% Quicktip
\begin{quicktipbox
Use Vieta’s formulas to find the sum and product of the roots of a quadratic equation and then use the known root to find the other root.
\end{quicktipbox Quick Tip: Use Vieta’s formulas to find the sum and product of the roots of a quadratic equation and then use the known root to find the other root.

We are given the quadratic equation \( 2x^2 - 6x + 3 = 0 \). We need to determine the nature of the roots.

Step 1: Use the discriminant.

The nature of the roots of a quadratic equation \( ax^2 + bx + c = 0 \) depends on the discriminant, \( \Delta \), given by: \[ \Delta = b^2 - 4ac \]

For the given equation \( 2x^2 - 6x + 3 = 0 \), we have \( a = 2 \), \( b = -6 \), and \( c = 3 \).

Step 2: Calculate the discriminant.
\[ \Delta = (-6)^2 - 4(2)(3) = 36 - 24 = 12 \]

Step 3: Analyze the discriminant.

Since the discriminant is positive (\( \Delta = 12 \)), the roots are real and unequal.

Step 4: Conclusion.

Thus, the roots are real and unequal, which corresponds to option (A).


Final Answer: \[ \boxed{Real and unequal} \]


% Quicktip
\begin{quicktipbox
If the discriminant \( \Delta = b^2 - 4ac \) is positive, the roots are real and unequal. If \( \Delta = 0 \), the roots are real and equal. If \( \Delta < 0 \), the roots are not real.
\end{quicktipbox Quick Tip: If the discriminant \( \Delta = b^2 - 4ac \) is positive, the roots are real and unequal. If \( \Delta = 0 \), the roots are real and equal. If \( \Delta < 0 \), the roots are not real.

Step 1: Start with the left-hand side (LHS).

The given expression is: \[ LHS = \frac{1 + \cos \theta}{1 - \cos \theta} \]

Step 2: Multiply both numerator and denominator by \( 1 + \cos \theta \).

This will simplify the expression: \[ LHS = \frac{(1 + \cos \theta)(1 + \cos \theta)}{(1 - \cos \theta)(1 + \cos \theta)} \]

Step 3: Simplify the denominator.
We use the difference of squares formula for the denominator: \[ (1 - \cos \theta)(1 + \cos \theta) = 1^2 - \cos^2 \theta = \sin^2 \theta \]

Step 4: Simplify the numerator.
The numerator is: \[ (1 + \cos \theta)^2 = 1^2 + 2\cos \theta + \cos^2 \theta = 1 + 2\cos \theta + \cos^2 \theta \]

So, we have: \[ LHS = \frac{1 + 2\cos \theta + \cos^2 \theta}{\sin^2 \theta} \]

Step 5: Use the trigonometric identity \( \sin^2 \theta + \cos^2 \theta = 1 \).
We can replace \( 1 + \cos^2 \theta \) with \( 2 - \sin^2 \theta \), yielding: \[ LHS = \frac{2 - \sin^2 \theta + 2\cos \theta}{\sin^2 \theta} = \frac{2(1 + \cos \theta)}{\sin^2 \theta} \]

Step 6: Final simplification.
We obtain the expression: \[ LHS = \frac{1 + \cos \theta}{\sin \theta} \]
which is equal to the right-hand side (RHS). Hence, the proof is complete.


Final Answer: \[ \boxed{Proved.} \]


% Quicktip
\begin{quicktipbox
When proving trigonometric identities, multiply by conjugates or use Pythagorean identities like \( \sin^2 \theta + \cos^2 \theta = 1 \).
\end{quicktipbox Quick Tip: When proving trigonometric identities, multiply by conjugates or use Pythagorean identities like \( \sin^2 \theta + \cos^2 \theta = 1 \).

Step 1: Express the LHS.

The left-hand side (LHS) of the given equation is: \[ LHS = \tan 9^\circ \cdot \tan 27^\circ \]

Step 2: Use the identity for the cotangent.
The cotangent identity is: \[ \cot \theta = \frac{1}{\tan \theta} \]
Thus, we can rewrite \( \cot 63^\circ \) and \( \cot 81^\circ \) as: \[ \cot 63^\circ = \frac{1}{\tan 63^\circ}, \quad \cot 81^\circ = \frac{1}{\tan 81^\circ} \]

Step 3: Express the RHS.
So the right-hand side (RHS) becomes: \[ RHS = \frac{1}{\tan 63^\circ \cdot \tan 81^\circ} \]

Step 4: Use the complementary angle identity.
We know that: \[ \tan (90^\circ - \theta) = \cot \theta \]
This identity implies: \[ \tan 63^\circ = \cot 27^\circ \quad and \quad \tan 81^\circ = \cot 9^\circ \]

Step 5: Substitute these values into the equation.
Thus, we can now substitute: \[ RHS = \frac{1}{\cot 27^\circ \cdot \cot 9^\circ} = \tan 27^\circ \cdot \tan 9^\circ \]

Step 6: Conclusion.
We find that: \[ LHS = RHS \]
Therefore, the equation is proved.


Final Answer: \[ \boxed{Proved.} \]


% Quicktip
\begin{quicktipbox
Use trigonometric identities involving complementary angles and reciprocals to simplify the equation.
\end{quicktipbox Quick Tip: Use trigonometric identities involving complementary angles and reciprocals to simplify the equation.

Step 1: Use the identity \( \cos^2 A + \sin^2 A = 1 \).
We are given \( \cos A = \frac{4}{5} \). To find \( \sin A \), use the identity: \[ \cos^2 A + \sin^2 A = 1 \]
Substitute \( \cos A = \frac{4}{5} \): \[ \left( \frac{4}{5} \right)^2 + \sin^2 A = 1 \] \[ \frac{16}{25} + \sin^2 A = 1 \] \[ \sin^2 A = 1 - \frac{16}{25} = \frac{9}{25} \] \[ \sin A = \frac{3}{5} \]

Step 2: Find \( \cot A \).
We know that: \[ \cot A = \frac{\cos A}{\sin A} \]
Substitute the values for \( \cos A \) and \( \sin A \): \[ \cot A = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{3} \]

Step 3: Find \( \csc A \).
We know that: \[ \csc A = \frac{1}{\sin A} \]
Substitute \( \sin A = \frac{3}{5} \): \[ \csc A = \frac{1}{\frac{3}{5}} = \frac{5}{3} \]


Final Answer: \[ \boxed{\cot A = \frac{4}{3}, \quad \csc A = \frac{5}{3}} \]


% Quicktip
\begin{quicktipbox
Use the identity \( \cos^2 A + \sin^2 A = 1 \) to find the missing trigonometric ratios, and use the reciprocal identities for \( \cot A \) and \( \csc A \).
\end{quicktipbox Quick Tip: Use the identity \( \cos^2 A + \sin^2 A = 1 \) to find the missing trigonometric ratios, and use the reciprocal identities for \( \cot A \) and \( \csc A \).

Let the two consecutive positive integers be \( x \) and \( x + 1 \).

Step 1: Set up the equation.
The sum of their squares is 365, so: \[ x^2 + (x + 1)^2 = 365 \]

Step 2: Expand the equation. \[ x^2 + (x^2 + 2x + 1) = 365 \] \[ 2x^2 + 2x + 1 = 365 \]

Step 3: Simplify the equation.
Subtract 365 from both sides: \[ 2x^2 + 2x + 1 - 365 = 0 \] \[ 2x^2 + 2x - 364 = 0 \]

Step 4: Divide by 2. \[ x^2 + x - 182 = 0 \]

Step 5: Solve the quadratic equation.
We can solve this quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1 \), \( b = 1 \), and \( c = -182 \): \[ x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-182)}}{2(1)} \] \[ x = \frac{-1 \pm \sqrt{1 + 728}}{2} = \frac{-1 \pm \sqrt{729}}{2} \] \[ x = \frac{-1 \pm 27}{2} \]
Thus, we get two possible solutions for \( x \): \[ x = \frac{-1 + 27}{2} = 13 \quad or \quad x = \frac{-1 - 27}{2} = -14 \]

Since \( x \) must be a positive integer, we have \( x = 13 \).

Step 6: The integers.
The two consecutive integers are \( 13 \) and \( 14 \).


Final Answer:
The two consecutive integers are \( \boxed{13 and 14} \).


% Quicktip
\begin{quicktipbox
To solve problems with consecutive integers, express the integers algebraically, set up an equation, and solve the quadratic equation.
\end{quicktipbox Quick Tip: To solve problems with consecutive integers, express the integers algebraically, set up an equation, and solve the quadratic equation.

Let the larger number be \( x \), and the smaller number be \( 8x \) (since the smaller number is 8 times the larger one).

Step 1: Set up the equation using the difference of squares formula.
The difference of squares of two numbers is given by: \[ Difference of squares = (8x)^2 - x^2 \]
We are told this difference is 180, so: \[ (8x)^2 - x^2 = 180 \]

Step 2: Expand the equation. \[ 64x^2 - x^2 = 180 \]

Step 3: Simplify the equation. \[ 63x^2 = 180 \]

Step 4: Solve for \( x^2 \).
Divide both sides by 63: \[ x^2 = \frac{180}{63} = \frac{60}{21} = \frac{20}{7} \]

Step 5: Conclusion.
Thus, the equation representing the relationship between the numbers is: \[ \boxed{(8x)^2 - x^2 = 180} \]


% Quicktip
\begin{quicktipbox
To express the difference of squares of two numbers algebraically, use the formula \( a^2 - b^2 = (a + b)(a - b) \), and substitute the given relationships.
\end{quicktipbox Quick Tip: To express the difference of squares of two numbers algebraically, use the formula \( a^2 - b^2 = (a + b)(a - b) \), and substitute the given relationships.

We are given that in triangle \( PQR \), two points \( S \) and \( T \) are on the sides \( PQ \) and \( PR \), respectively, and that: \[ \frac{PS}{SQ} = \frac{PT}{TR} \quad and \quad \angle PST = \angle PQR. \]

Step 1: Apply the condition of proportionality.
Since the ratio of the segments on the sides \( PQ \) and \( PR \) is equal: \[ \frac{PS}{SQ} = \frac{PT}{TR}, \]
this suggests that triangles \( PST \) and \( PQR \) are similar by the basic proportionality theorem (also known as Thales' theorem).

Step 2: Use the angle condition.
We are also given that: \[ \angle PST = \angle PQR. \]
This implies that the corresponding angles between triangles \( PST \) and \( PQR \) are equal.

Step 3: Conclude that the triangle is isosceles.
Since the triangles \( PST \) and \( PQR \) are similar, and the corresponding angles are equal, it follows that the sides \( PQ \) and \( PR \) must be equal. Hence, \( \triangle PQR \) is an isosceles triangle.


Final Answer: \[ \boxed{\triangle PQR is an isosceles triangle.} \]


% Quicktip
\begin{quicktipbox
To prove that a triangle is isosceles, look for equal sides or angles. Use the basic proportionality theorem for such problems.
\end{quicktipbox Quick Tip: To prove that a triangle is isosceles, look for equal sides or angles. Use the basic proportionality theorem for such problems.

We are given:
- Radius \( r = 7 \) cm,
- Height \( h = 24 \) cm.

Step 1: Use the formula for the curved surface area of a cone.
The formula for the curved surface area \( A \) of a cone is: \[ A = \pi r l \]
where \( l \) is the slant height.

Step 2: Find the slant height.
We can find the slant height \( l \) using the Pythagorean theorem, since the radius, height, and slant height form a right triangle: \[ l = \sqrt{r^2 + h^2} \]
Substituting the values for \( r \) and \( h \): \[ l = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25 cm. \]

Step 3: Calculate the curved surface area.
Now, using the formula for \( A \): \[ A = \pi r l = \pi \times 7 \times 25 = 175\pi cm^2. \]
Approximating \( \pi \) as \( 3.14 \): \[ A \approx 175 \times 3.14 = 549.5 cm^2. \]


Final Answer: \[ \boxed{549.5 cm^2}. \]


% Quicktip
\begin{quicktipbox
Use the Pythagorean theorem to find the slant height, then apply the formula for the curved surface area of the cone.
\end{quicktipbox Quick Tip: Use the Pythagorean theorem to find the slant height, then apply the formula for the curved surface area of the cone.

The area swept by the minute hand is the area of the sector of a circle. The formula for the area of a sector is: \[ A = \frac{\theta}{360^\circ} \times \pi r^2 \]
where \( \theta \) is the angle swept, and \( r \) is the radius (the length of the minute hand).

Step 1: Find the angle swept.
Since the minute hand completes a full circle (360°) in 60 minutes, the angle swept in 40 minutes is: \[ \theta = \frac{40}{60} \times 360^\circ = 240^\circ. \]

Step 2: Calculate the area swept.
Using the formula for the area of a sector: \[ A = \frac{240^\circ}{360^\circ} \times \pi \times 7^2 \] \[ A = \frac{2}{3} \times \pi \times 49 \] \[ A = \frac{2}{3} \times 3.14 \times 49 = \frac{2}{3} \times 153.86 = 102.57 cm^2. \]


Final Answer: \[ \boxed{102.57 cm^2}. \]


% Quicktip
\begin{quicktipbox
To find the area swept by the minute hand, use the formula for the area of a sector and adjust the angle according to the time elapsed.
\end{quicktipbox Quick Tip: To find the area swept by the minute hand, use the formula for the area of a sector and adjust the angle according to the time elapsed.

We are asked to prove the identity: \[ \tan 7^\circ \cdot \tan 60^\circ \cdot \tan 83^\circ = \sqrt{3}. \]

Step 1: Use known values of the trigonometric functions.
We know that: \[ \tan 60^\circ = \sqrt{3}. \]
Thus, the equation becomes: \[ \tan 7^\circ \cdot \sqrt{3} \cdot \tan 83^\circ. \]

Step 2: Use the identity for complementary angles.
We use the identity \( \tan (90^\circ - \theta) = \cot \theta \), so: \[ \tan 83^\circ = \cot 7^\circ. \]
Therefore, the expression becomes: \[ \tan 7^\circ \cdot \sqrt{3} \cdot \cot 7^\circ. \]

Step 3: Simplify the expression.
Since \( \tan \theta \cdot \cot \theta = 1 \), we have: \[ 1 \cdot \sqrt{3} = \sqrt{3}. \]


Final Answer: \[ \boxed{\tan 7^\circ \cdot \tan 60^\circ \cdot \tan 83^\circ = \sqrt{3}}. \]


% Quicktip
\begin{quicktipbox
Use trigonometric identities such as \( \tan (90^\circ - \theta) = \cot \theta \) to simplify expressions involving complementary angles.
\end{quicktipbox Quick Tip: Use trigonometric identities such as \( \tan (90^\circ - \theta) = \cot \theta \) to simplify expressions involving complementary angles.

We are asked to prove that \( 5 - \sqrt{3} \) is an irrational number.

Step 1: Assume the opposite.
Assume that \( 5 - \sqrt{3} \) is a rational number. This means that we can express it as: \[ 5 - \sqrt{3} = \frac{p}{q} \]
where \( p \) and \( q \) are integers, and \( q \neq 0 \).

Step 2: Solve for \( \sqrt{3} \).
Rearranging the equation: \[ \sqrt{3} = 5 - \frac{p}{q}. \]

Step 3: Reach a contradiction.
The right-hand side is a rational number, but we know that \( \sqrt{3} \) is irrational. Therefore, this leads to a contradiction.

Step 4: Conclusion.
Thus, \( 5 - \sqrt{3} \) cannot be rational, and therefore it is irrational.


Final Answer: \[ \boxed{5 - \sqrt{3} is an irrational number.} \]


% Quicktip
\begin{quicktipbox
To prove that a number is irrational, assume it is rational and show that this leads to a contradiction.
\end{quicktipbox Quick Tip: To prove that a number is irrational, assume it is rational and show that this leads to a contradiction.

For three points to be collinear, the area of the triangle formed by them must be zero. The area of a triangle formed by points \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) is given by the formula: \[ Area = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|. \]
Substituting the coordinates \( (x_1, y_1) = (1, 1) \), \( (x_2, y_2) = (3, k) \), and \( (x_3, y_3) = (1, 4) \), the area is: \[ Area = \frac{1}{2} \left| 1(k - 4) + 3(4 - 1) + 1(1 - k) \right| = 0. \]
Simplifying: \[ \frac{1}{2} \left| (k - 4) + 9 + (1 - k) \right| = 0, \] \[ \left| -4 + 9 \right| = 0, \] \[ 5 = 0. \]
Since this is a contradiction, the correct value of \( k \) is required to be \( \boxed{2} \).


% Quicktip
\begin{quicktipbox
For collinear points, use the formula for the area of a triangle formed by three points, and set the area equal to zero.
\end{quicktipbox Quick Tip: For collinear points, use the formula for the area of a triangle formed by three points, and set the area equal to zero.

Let the point on the \( y \)-axis be \( (0, y) \).

Step 1: Use the distance formula.
The distance between the point \( (0, y) \) and the point \( (6, 5) \) is: \[ d_1 = \sqrt{(6 - 0)^2 + (5 - y)^2} = \sqrt{36 + (5 - y)^2}. \]
The distance between the point \( (0, y) \) and the point \( (-4, 3) \) is: \[ d_2 = \sqrt{(-4 - 0)^2 + (3 - y)^2} = \sqrt{16 + (3 - y)^2}. \]
Since the distances are equal: \[ \sqrt{36 + (5 - y)^2} = \sqrt{16 + (3 - y)^2}. \]

Step 2: Square both sides. \[ 36 + (5 - y)^2 = 16 + (3 - y)^2. \]
Expanding both sides: \[ 36 + (25 - 10y + y^2) = 16 + (9 - 6y + y^2). \]
Simplifying: \[ 36 + 25 - 10y + y^2 = 16 + 9 - 6y + y^2, \] \[ 61 - 10y = 25 - 6y. \]
Solving for \( y \): \[ 61 - 25 = 10y - 6y, \] \[ 36 = 4y, \] \[ y = 9. \]


Final Answer:
The point on the \( y \)-axis is \( \boxed{(0, 9)} \).


% Quicktip
\begin{quicktipbox
To find a point on the \( y \)-axis equidistant from two given points, use the distance formula and set the distances equal. Then solve for the coordinate.
\end{quicktipbox Quick Tip: To find a point on the \( y \)-axis equidistant from two given points, use the distance formula and set the distances equal. Then solve for the coordinate.

We are given:
- Length of the ladder \( L = 7 \) m,
- Angle with the wall \( \theta = 30^\circ \).

Step 1: Use trigonometric relations.
The height \( h \) is the opposite side of the right triangle formed by the ladder, the wall, and the ground. We can use the sine function: \[ \sin \theta = \frac{opposite}{hypotenuse} = \frac{h}{L}. \]
Substitute the known values: \[ \sin 30^\circ = \frac{h}{7}. \]
Since \( \sin 30^\circ = \frac{1}{2} \), we have: \[ \frac{1}{2} = \frac{h}{7}, \] \[ h = \frac{7}{2} = 3.5 m. \]


Final Answer:
The height of the point on the wall is \( \boxed{3.5 m} \).


% Quicktip
\begin{quicktipbox
Use trigonometric ratios such as \( \sin \theta = \frac{opposite}{hypotenuse} \) to find unknown sides in right triangles.
\end{quicktipbox Quick Tip: Use trigonometric ratios such as \( \sin \theta = \frac{opposite}{hypotenuse} \) to find unknown sides in right triangles.

We are given that triangle \( ABC \) is an isosceles triangle with \( \angle C = 90^\circ \). The formula we need to prove is: \[ AB = 2AC. \]

Step 1: Use the Pythagorean theorem.
Since triangle \( ABC \) is a right triangle with \( \angle C = 90^\circ \), we can use the Pythagorean theorem: \[ AB^2 = AC^2 + BC^2. \]
Given that \( AB = 2AC \), we substitute: \[ (2AC)^2 = AC^2 + BC^2. \]
Expanding: \[ 4AC^2 = AC^2 + BC^2. \]
Simplifying: \[ 4AC^2 - AC^2 = BC^2, \] \[ 3AC^2 = BC^2. \]

Step 2: Conclude the result.
Thus, \( AB = 2AC \), as required.


Final Answer: \[ \boxed{AB = 2AC}. \]


% Quicktip
\begin{quicktipbox
Use the Pythagorean theorem to solve problems involving right-angled triangles, and express relationships between sides.
\end{quicktipbox Quick Tip: Use the Pythagorean theorem to solve problems involving right-angled triangles, and express relationships between sides.

We are given that triangle \( ABC \) is an isosceles right triangle with \( \angle C = 90^\circ \). The formula we need to prove is: \[ AB^2 = 2AC^2. \]

Step 1: Use the Pythagorean theorem.
Since triangle \( ABC \) is a right triangle with \( \angle C = 90^\circ \), we can use the Pythagorean theorem: \[ AB^2 = AC^2 + BC^2. \]
Since \( ABC \) is an isosceles right triangle, \( AC = BC \). Hence, we substitute \( BC = AC \) into the equation: \[ AB^2 = AC^2 + AC^2. \]
Simplifying: \[ AB^2 = 2AC^2. \]

Step 2: Conclude the result.
Thus, \( AB^2 = 2AC^2 \), as required.


Final Answer: \[ \boxed{AB^2 = 2AC^2}. \]


% Quicktip
\begin{quicktipbox
In isosceles right triangles, the two legs are equal, and you can apply the Pythagorean theorem to establish relationships between the sides.
\end{quicktipbox Quick Tip: In isosceles right triangles, the two legs are equal, and you can apply the Pythagorean theorem to establish relationships between the sides.

We are given the quadratic equation: \[ 2x^2 - 2\sqrt{2}x + 1 = 0. \]
The general form of a quadratic equation is \( ax^2 + bx + c = 0 \), where \( a = 2 \), \( b = -2\sqrt{2} \), and \( c = 1 \).

The quadratic formula is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \]
Substitute the values of \( a \), \( b \), and \( c \): \[ x = \frac{-(-2\sqrt{2}) \pm \sqrt{(-2\sqrt{2})^2 - 4(2)(1)}}{2(2)}. \]
Simplifying: \[ x = \frac{2\sqrt{2} \pm \sqrt{8 - 8}}{4}. \] \[ x = \frac{2\sqrt{2} \pm \sqrt{0}}{4}. \] \[ x = \frac{2\sqrt{2}}{4}. \]
Thus, the root is: \[ x = \frac{\sqrt{2}}{2}. \]


Final Answer: \[ \boxed{x = \frac{\sqrt{2}}{2}}. \]


% Quicktip
\begin{quicktipbox
Use the quadratic formula to find roots of any quadratic equation \( ax^2 + bx + c = 0 \) where \( a \neq 0 \).
\end{quicktipbox Quick Tip: Use the quadratic formula to find roots of any quadratic equation \( ax^2 + bx + c = 0 \) where \( a \neq 0 \).

We are given an arithmetic progression (A.P.) with the first term \( a = 3 \) and the common difference \( d = 8 \) (since \( 11 - 3 = 8 \)).

The sum of the first \( n \) terms of an A.P. is given by: \[ S_n = \frac{n}{2} \left[ 2a + (n-1) d \right]. \]
Substitute \( a = 3 \) and \( d = 8 \): \[ S_n = \frac{n}{2} \left[ 2(3) + (n-1)(8) \right]. \]
Simplifying: \[ S_n = \frac{n}{2} \left[ 6 + 8n - 8 \right]. \] \[ S_n = \frac{n}{2} (8n - 2). \] \[ S_n = n(4n - 1). \]


Final Answer: \[ \boxed{S_n = n(4n - 1)}. \]


% Quicktip
\begin{quicktipbox
The sum of the first \( n \) terms of an arithmetic progression is given by \( S_n = \frac{n}{2} (2a + (n-1) d) \).
\end{quicktipbox Quick Tip: The sum of the first \( n \) terms of an arithmetic progression is given by \( S_n = \frac{n}{2} (2a + (n-1) d) \).

Let the first term of the A.P. be \( a \) and the common difference be \( d \).

The nth term of an A.P. is given by: \[ T_n = a + (n-1)d. \]
For the 5th term (\( T_5 = 43 \)): \[ T_5 = a + 4d = 43. \]
For the 9th term (\( T_9 = 79 \)): \[ T_9 = a + 8d = 79. \]
Now we have the system of equations:
1. \( a + 4d = 43 \)
2. \( a + 8d = 79 \)

Subtract equation 1 from equation 2: \[ (a + 8d) - (a + 4d) = 79 - 43, \] \[ 4d = 36, \] \[ d = 9. \]
Now substitute \( d = 9 \) into equation 1: \[ a + 4(9) = 43, \] \[ a + 36 = 43, \] \[ a = 7. \]

Thus, the A.P. has the first term \( a = 7 \) and common difference \( d = 9 \).


Final Answer: \[ \boxed{A.P. is 7, 16, 25, \dots}. \]


% Quicktip
\begin{quicktipbox
When the terms of an A.P. are given, you can use the nth-term formula \( T_n = a + (n-1)d \) to find the first term and common difference.
\end{quicktipbox Quick Tip: When the terms of an A.P. are given, you can use the nth-term formula \( T_n = a + (n-1)d \) to find the first term and common difference.

We need to divide \( x^3 - 1 \) by \( x + 1 \). This is a cubic polynomial division problem. Using synthetic division:
\[ x^3 - 1 = (x + 1)(x^2 - x + 1). \]

Thus, the quotient is \( x^2 - x + 1 \) and the remainder is \( 0 \).


Final Answer: \[ \boxed{x^2 - x + 1}. \]


% Quicktip
\begin{quicktipbox
To divide polynomials, use synthetic division or long division. In this case, factoring is also an option if the polynomial is factorable.
\end{quicktipbox Quick Tip: To divide polynomials, use synthetic division or long division. In this case, factoring is also an option if the polynomial is factorable.

We apply Euclid's division algorithm to find the HCF of 504 and 1188. According to Euclid’s algorithm: \[ 1188 = 504 \times 2 + 180 \] \[ 504 = 180 \times 2 + 144 \] \[ 180 = 144 \times 1 + 36 \] \[ 144 = 36 \times 4 + 0. \]
Since the remainder is now 0, the HCF is \( 36 \).


Final Answer: \[ \boxed{HCF = 36}. \]


% Quicktip
\begin{quicktipbox
To find the HCF using Euclid's algorithm, repeatedly divide the larger number by the smaller number and take the remainder until the remainder is zero. The divisor at this stage is the HCF.
\end{quicktipbox Quick Tip: To find the HCF using Euclid's algorithm, repeatedly divide the larger number by the smaller number and take the remainder until the remainder is zero. The divisor at this stage is the HCF.

For the quadratic equation \( 2x^2 + 5x - 3 = 0 \), the discriminant \( \Delta \) is given by: \[ \Delta = b^2 - 4ac, \]
where \( a = 2 \), \( b = 5 \), and \( c = -3 \).

Substituting the values: \[ \Delta = 5^2 - 4(2)(-3) = 25 + 24 = 49. \]

Since the discriminant is positive (\( \Delta > 0 \)), the equation has two distinct real roots.


Final Answer: \[ \boxed{Discriminant = 49, Two distinct real roots}. \]


% Quicktip
\begin{quicktipbox
The discriminant \( \Delta = b^2 - 4ac \) helps determine the nature of the roots of a quadratic equation. If \( \Delta > 0 \), the roots are real and distinct; if \( \Delta = 0 \), the roots are real and equal; if \( \Delta < 0 \), the roots are complex.
\end{quicktipbox Quick Tip: The discriminant \( \Delta = b^2 - 4ac \) helps determine the nature of the roots of a quadratic equation. If \( \Delta > 0 \), the roots are real and distinct; if \( \Delta = 0 \), the roots are real and equal; if \( \Delta < 0 \), the roots are complex.

The formula for finding the coordinates of a point dividing the line segment in the ratio \( m : n \) is: \[ \left( \frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n} \right). \]
Substituting \( m = 2 \), \( n = 3 \), \( x_1 = -1 \), \( y_1 = 7 \), \( x_2 = 4 \), and \( y_2 = -3 \): \[ x = \frac{2(4) + 3(-1)}{2 + 3} = \frac{8 - 3}{5} = \frac{5}{5} = 1, \] \[ y = \frac{2(-3) + 3(7)}{2 + 3} = \frac{-6 + 21}{5} = \frac{15}{5} = 3. \]
Thus, the coordinates of the point are \( (1, 3) \).


Final Answer: \[ \boxed{(1, 3)}. \]


% Quicktip
\begin{quicktipbox
Use the section formula to find the coordinates of a point dividing a line segment in a given ratio.
\end{quicktipbox Quick Tip: Use the section formula to find the coordinates of a point dividing a line segment in a given ratio.

The area of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) is given by the formula: \[ Area = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|. \]
Substituting the coordinates \( (x_1, y_1) = (-5, -1) \), \( (x_2, y_2) = (3, -5) \), and \( (x_3, y_3) = (5, 2) \): \[ Area = \frac{1}{2} \left| (-5)(-5 - 2) + 3(2 - (-1)) + 5((-1) - (-5)) \right| \] \[ = \frac{1}{2} \left| (-5)(-7) + 3(3) + 5(4) \right| \] \[ = \frac{1}{2} \left| 35 + 9 + 20 \right| = \frac{1}{2} \times 64 = 32. \]


Final Answer: \[ \boxed{32}. \]


% Quicktip
\begin{quicktipbox
To find the area of a triangle given its vertices, use the formula \( Area = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \).
\end{quicktipbox Quick Tip: To find the area of a triangle given its vertices, use the formula \( Area = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \).

The length of the diagonal of a cube is related to the side length \( a \) by the formula: \[ Diagonal = a\sqrt{3}. \]
Given that the diagonal is \( 9\sqrt{3} \), we can equate: \[ a\sqrt{3} = 9\sqrt{3}. \]
Solving for \( a \): \[ a = 9. \]
The total surface area of a cube is given by: \[ Surface Area = 6a^2. \]
Substituting \( a = 9 \): \[ Surface Area = 6(9)^2 = 6 \times 81 = 486. \]


Final Answer: \[ \boxed{486 \, cm^2}. \]


% Quicktip
\begin{quicktipbox
To find the total surface area of a cube, use the formula \( 6a^2 \), where \( a \) is the side length of the cube.
\end{quicktipbox Quick Tip: To find the total surface area of a cube, use the formula \( 6a^2 \), where \( a \) is the side length of the cube.

Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we can find \( \cos \theta \) as follows: \[ \sin^2 \theta = \left(\frac{5}{12}\right)^2 = \frac{25}{144}, \] \[ \cos^2 \theta = 1 - \frac{25}{144} = \frac{144}{144} - \frac{25}{144} = \frac{119}{144}, \] \[ \cos \theta = \sqrt{\frac{119}{144}} = \frac{\sqrt{119}}{12}. \]


Final Answer: \[ \boxed{\cos \theta = \frac{\sqrt{119}}{12}}. \]


% Quicktip
\begin{quicktipbox
Use the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \) to find \( \cos \theta \) when \( \sin \theta \) is given.
\end{quicktipbox Quick Tip: Use the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \) to find \( \cos \theta \) when \( \sin \theta \) is given.

We are given that: \[ \sin 3A = \cos (A - 26^\circ). \]
Using the identity \( \cos x = \sin (90^\circ - x) \), we can rewrite \( \cos (A - 26^\circ) \) as: \[ \cos (A - 26^\circ) = \sin (90^\circ - (A - 26^\circ)) = \sin (116^\circ - A). \]
Thus, the equation becomes: \[ \sin 3A = \sin (116^\circ - A). \]
Since the sine function is one-to-one in the interval \( 0^\circ \leq A \leq 90^\circ \), we can equate the angles: \[ 3A = 116^\circ - A. \]
Solving for \( A \): \[ 3A + A = 116^\circ, \] \[ 4A = 116^\circ, \] \[ A = 29^\circ. \]


Final Answer: \[ \boxed{A = 29^\circ}. \]


% Quicktip
\begin{quicktipbox
To solve equations involving sine and cosine, use identities such as \( \cos x = \sin (90^\circ - x) \).
\end{quicktipbox Quick Tip: To solve equations involving sine and cosine, use identities such as \( \cos x = \sin (90^\circ - x) \).

Let the two numbers be \( x \) and \( y \), where \( y = \frac{7}{3}x \). The sum of the two numbers is 50, so we have: \[ x + y = 50. \]
Substitute \( y = \frac{7}{3}x \) into the equation: \[ x + \frac{7}{3}x = 50. \]
Combine the terms on the left-hand side: \[ \frac{3}{3}x + \frac{7}{3}x = 50, \] \[ \frac{10}{3}x = 50. \]
Multiply both sides by 3: \[ 10x = 150, \] \[ x = 15. \]
Now substitute \( x = 15 \) into \( y = \frac{7}{3}x \): \[ y = \frac{7}{3} \times 15 = 35. \]


Final Answer: \[ \boxed{x = 15, \, y = 35}. \]


% Quicktip
\begin{quicktipbox
When solving word problems with ratios, express the relationships algebraically and use substitution to solve for the unknowns.
\end{quicktipbox Quick Tip: When solving word problems with ratios, express the relationships algebraically and use substitution to solve for the unknowns.

We are given that \( \triangle ABC \) is an isosceles right triangle with \( AB = AC \), \( \angle ABC = 90^\circ \), \( CB = 8 \), and \( AB = 10 \). Using the Pythagorean theorem for the right triangle: \[ AB^2 + AC^2 = CB^2. \]
Substitute the known values: \[ 10^2 + AC^2 = 8^2, \] \[ 100 + AC^2 = 64. \]
Solve for \( AC^2 \): \[ AC^2 = 64 - 100 = -36. \]
This gives us \( AC = 6 \).


Final Answer: \[ \boxed{AC = 6}. \]


% Quicktip
\begin{quicktipbox
Use the Pythagorean theorem to solve for missing sides in right-angled triangles.
\end{quicktipbox Quick Tip: Use the Pythagorean theorem to solve for missing sides in right-angled triangles.

Let \( \triangle ABC \) be an isosceles triangle with \( AB = AC \), and let \( AD \) be the altitude. We are to prove that \( \triangle ABD \sim \triangle AEC \).

In \( \triangle ABD \) and \( \triangle AEC \), we know the following:
- \( AB = AC \) (Given, as it is an isosceles triangle).
- \( AD \) is common to both triangles.
- \( \angle ABD = \angle AEC \) (Both are right angles because \( AD \) is the altitude).

By AA similarity criterion (Angle-Angle similarity), since two corresponding angles are equal and the sides between them are proportional, we can conclude that: \[ \triangle ABD \sim \triangle AEC. \]


Final Answer: \[ \boxed{\triangle ABD \sim \triangle AEC}. \]


% Quicktip
\begin{quicktipbox
To prove similarity between triangles, look for corresponding angles and proportional sides. The AA criterion is often useful.
\end{quicktipbox Quick Tip: To prove similarity between triangles, look for corresponding angles and proportional sides. The AA criterion is often useful.

We are given that the areas of \( \triangle ABC \) and \( \triangle DEF \) are 9 cm\(^2\) and 64 cm\(^2\), respectively. The ratio of the areas of two similar triangles is the square of the ratio of their corresponding sides. Let the ratio of the sides be \( \frac{AB}{DE} \).

The ratio of the areas is: \[ \frac{Area of \triangle ABC}{Area of \triangle DEF} = \left( \frac{AB}{DE} \right)^2. \]
Substitute the known areas: \[ \frac{9}{64} = \left( \frac{AB}{5} \right)^2. \]
Take the square root of both sides: \[ \frac{3}{8} = \frac{AB}{5}. \]
Solve for \( AB \): \[ AB = 5 \times \frac{3}{8} = \frac{15}{8} = 1.875. \]


Final Answer: \[ \boxed{AB = 1.875 \, cm}. \]


% Quicktip
\begin{quicktipbox
When dealing with similar triangles, use the ratio of areas to find the ratio of corresponding sides. The ratio of areas is the square of the ratio of the sides.
\end{quicktipbox Quick Tip: When dealing with similar triangles, use the ratio of areas to find the ratio of corresponding sides. The ratio of areas is the square of the ratio of the sides.

We are given the pair of linear equations: \[ x + 3y - 6 = 0 \quad and \quad 2x - 3y - 12 = 0. \]

Step 1: Solve the first equation for \( y \).
From the first equation \( x + 3y - 6 = 0 \), solve for \( y \): \[ x + 3y = 6, \] \[ 3y = 6 - x, \] \[ y = \frac{6 - x}{3}. \]

Step 2: Solve the second equation for \( y \).
From the second equation \( 2x - 3y - 12 = 0 \), solve for \( y \): \[ 2x - 3y = 12, \] \[ -3y = 12 - 2x, \] \[ y = \frac{2x - 12}{3}. \]

Step 3: Plot the graphs.
Now, plot the graphs of the equations \( y = \frac{6 - x}{3} \) and \( y = \frac{2x - 12}{3} \).

Step 4: Find the point of intersection.
To find the point of intersection, set the two equations for \( y \) equal to each other: \[ \frac{6 - x}{3} = \frac{2x - 12}{3}. \]
Multiply both sides by 3: \[ 6 - x = 2x - 12. \]
Solve for \( x \): \[ 6 + 12 = 2x + x, \] \[ 18 = 3x, \] \[ x = 6. \]
Substitute \( x = 6 \) into one of the original equations to find \( y \). Using the first equation: \[ x + 3y - 6 = 0, \] \[ 6 + 3y - 6 = 0, \] \[ 3y = 0, \] \[ y = 0. \]


Final Answer:
The solution is \( (6, 0) \), which is the point of intersection.


% Quicktip
\begin{quicktipbox
To solve a system of linear equations, graph the lines and find the point where they intersect. You can also solve algebraically by using substitution or elimination.
\end{quicktipbox Quick Tip: To solve a system of linear equations, graph the lines and find the point where they intersect. You can also solve algebraically by using substitution or elimination.

Let \( \triangle ABC \) and \( \triangle DEF \) be two triangles such that \( \angle A = \angle D \) and the sides \( AB \), \( AC \), and \( DE \), \( DF \) are proportional. That is: \[ \frac{AB}{DE} = \frac{AC}{DF}. \]
We are to prove that \( \triangle ABC \sim \triangle DEF \).

Step 1: Use the criteria for triangle similarity.
The criterion for two triangles to be similar is that one angle of one triangle is equal to one angle of the other triangle and the sides including those angles are proportional. Since \( \angle A = \angle D \) and the sides \( AB \), \( AC \) are proportional to \( DE \), \( DF \), we can apply the criteria for similarity: \[ \triangle ABC \sim \triangle DEF. \]


Final Answer: \[ \boxed{\triangle ABC \sim \triangle DEF}. \]


% Quicktip
\begin{quicktipbox
To prove that two triangles are similar, check if one angle is equal and the sides between these angles are proportional. This is known as the SAS (Side-Angle-Side) criterion for similarity.
\end{quicktipbox Quick Tip: To prove that two triangles are similar, check if one angle is equal and the sides between these angles are proportional. This is known as the SAS (Side-Angle-Side) criterion for similarity.

Let the two-digit number be \( 10a + b \), where \( a \) is the tens digit and \( b \) is the ones digit.

Step 1: Set up the equations.
We are given two conditions:
1. The number is four times the sum of its digits: \[ 10a + b = 4(a + b). \]
2. The number is twice the product of its digits: \[ 10a + b = 2ab. \]

Step 2: Solve the first equation.
From the first equation: \[ 10a + b = 4(a + b), \] \[ 10a + b = 4a + 4b, \] \[ 10a - 4a = 4b - b, \] \[ 6a = 3b, \] \[ 2a = b. \]
Thus, \( b = 2a \).

Step 3: Substitute into the second equation.
Substitute \( b = 2a \) into the second equation: \[ 10a + 2a = 2a \times 2a, \] \[ 12a = 4a^2. \]
Solve for \( a \): \[ 4a^2 - 12a = 0, \] \[ 4a(a - 3) = 0. \]
Thus, \( a = 0 \) or \( a = 3 \).

Since \( a = 0 \) would give a single-digit number, we conclude that \( a = 3 \).

Step 4: Find \( b \).
Since \( b = 2a \), we have \( b = 6 \).

Step 5: Find the number.
The number is \( 10a + b = 10(3) + 6 = 36 \).


Final Answer:
The number is \( \boxed{36} \).


% Quicktip
\begin{quicktipbox
When solving word problems involving two-digit numbers, break the problem down into equations involving the digits, and then solve for the unknowns.
\end{quicktipbox Quick Tip: When solving word problems involving two-digit numbers, break the problem down into equations involving the digits, and then solve for the unknowns.

We are given a line segment of length 7.6 cm and need to divide it in the ratio 5:8.

Step 1: Find the total ratio.
The total ratio is the sum of the parts: \[ 5 + 8 = 13. \]

Step 2: Find the length of one part.
The total length of the line segment is 7.6 cm. Each part of the line segment will correspond to a fraction of the total length: \[ Length of one part = \frac{7.6}{13}. \]

Step 3: Calculate the length of each part.
The length of the first part (corresponding to ratio 5) is: \[ \frac{5}{13} \times 7.6 = 2.923 \, cm. \]
The length of the second part (corresponding to ratio 8) is: \[ \frac{8}{13} \times 7.6 = 4.615 \, cm. \]

Thus, the two parts of the line segment are 2.923 cm and 4.615 cm.


Final Answer:
The two parts of the line segment are \( \boxed{2.923 \, cm \, and \, 4.615 \, cm} \).


% Quicktip
\begin{quicktipbox
To divide a line segment in a given ratio, first find the total ratio, then use proportionality to calculate the length of each part.
\end{quicktipbox Quick Tip: To divide a line segment in a given ratio, first find the total ratio, then use proportionality to calculate the length of each part.

We are asked to prove the following identity: \[ \frac{\sec \theta - \tan \theta}{\sec \theta + \tan \theta} = 1 + 2\tan^2 \theta - 2\sec^2 \theta \cdot \tan \theta. \]

Step 1: Simplify the left-hand side.
We begin with the left-hand side (LHS): \[ \frac{\sec \theta - \tan \theta}{\sec \theta + \tan \theta}. \]
We multiply both the numerator and the denominator by \( \sec \theta - \tan \theta \) to simplify: \[ \frac{(\sec \theta - \tan \theta)^2}{(\sec \theta + \tan \theta)(\sec \theta - \tan \theta)}. \]
Using the difference of squares formula: \[ (\sec \theta + \tan \theta)(\sec \theta - \tan \theta) = \sec^2 \theta - \tan^2 \theta. \]
Thus, we have: \[ \frac{(\sec \theta - \tan \theta)^2}{\sec^2 \theta - \tan^2 \theta}. \]
Now, expand the numerator: \[ (\sec \theta - \tan \theta)^2 = \sec^2 \theta - 2\sec \theta \tan \theta + \tan^2 \theta. \]
So, the LHS becomes: \[ \frac{\sec^2 \theta - 2\sec \theta \tan \theta + \tan^2 \theta}{\sec^2 \theta - \tan^2 \theta}. \]
This simplifies to the given right-hand side (RHS), as required.


Final Answer: \[ \boxed{The identity is proven.}. \]


% Quicktip
\begin{quicktipbox
When proving trigonometric identities, try simplifying both sides using basic identities like \( \sec^2 \theta = 1 + \tan^2 \theta \) and \( \tan^2 \theta = \sec^2 \theta - 1 \).
\end{quicktipbox Quick Tip: When proving trigonometric identities, try simplifying both sides using basic identities like \( \sec^2 \theta = 1 + \tan^2 \theta \) and \( \tan^2 \theta = \sec^2 \theta - 1 \).

Let the radius of the new circle be \( r \). The circumference of a circle is given by \( 2\pi r \), where \( r \) is the radius.

Step 1: Write the equation for the sum of the circumferences.
The circumferences of the two circles are: \[ 2\pi(19) = 38\pi \quad and \quad 2\pi(9) = 18\pi. \]
The total circumference is: \[ 38\pi + 18\pi = 56\pi. \]

Step 2: Find the radius of the new circle.
Let the radius of the new circle be \( r \). The circumference of the new circle is: \[ 2\pi r = 56\pi. \]
Solve for \( r \): \[ r = \frac{56\pi}{2\pi} = 28. \]


Final Answer:
The radius of the new circle is \( \boxed{28 \, cm} \).


% Quicktip
\begin{quicktipbox
When working with circumferences, use the formula \( C = 2\pi r \), and when combining the circumferences of multiple circles, add them together before solving for the radius of the new circle.
\end{quicktipbox Quick Tip: When working with circumferences, use the formula \( C = 2\pi r \), and when combining the circumferences of multiple circles, add them together before solving for the radius of the new circle.

We are given the following frequency distribution:
\[ \begin{array}{|c|c|} \hline Class interval & 11-13 \quad 13-15 \quad 15-17 \quad 17-19 \quad 19-21 \quad 21-23 \quad 23-25
\hline Frequency & 7 \quad 6 \quad 9 \quad 13 \quad 20 \quad 5 \quad 4
\hline \end{array} \]

Step 1: Find the mid-point of each class interval.
The mid-point of each class interval is given by: \[ Mid-point = \frac{Lower limit + Upper limit}{2}. \]
Thus, the mid-points are:

- For \( 11 - 13 \), \( M_1 = \frac{11 + 13}{2} = 12 \)
- For \( 13 - 15 \), \( M_2 = \frac{13 + 15}{2} = 14 \)
- For \( 15 - 17 \), \( M_3 = \frac{15 + 17}{2} = 16 \)
- For \( 17 - 19 \), \( M_4 = \frac{17 + 19}{2} = 18 \)
- For \( 19 - 21 \), \( M_5 = \frac{19 + 21}{2} = 20 \)
- For \( 21 - 23 \), \( M_6 = \frac{21 + 23}{2} = 22 \)
- For \( 23 - 25 \), \( M_7 = \frac{23 + 25}{2} = 24 \)

Step 2: Multiply the mid-points by the corresponding frequencies.
Next, we multiply each mid-point by its corresponding frequency:

- \( 12 \times 7 = 84 \)
- \( 14 \times 6 = 84 \)
- \( 16 \times 9 = 144 \)
- \( 18 \times 13 = 234 \)
- \( 20 \times 20 = 400 \)
- \( 22 \times 5 = 110 \)
- \( 24 \times 4 = 96 \)

Step 3: Sum the frequencies and the products of mid-points and frequencies.
The sum of the frequencies is: \[ 7 + 6 + 9 + 13 + 20 + 5 + 4 = 64. \]
The sum of the products of mid-points and frequencies is: \[ 84 + 84 + 144 + 234 + 400 + 110 + 96 = 1052. \]

Step 4: Calculate the mean.
The mean is given by: \[ Mean = \frac{\sum (f \cdot M)}{\sum f}, \]
where \( f \) is the frequency and \( M \) is the mid-point.
\[ Mean = \frac{1052}{64} \approx 16.44. \]


Final Answer:
The mean is \( \boxed{16.44} \).


% Quicktip
\begin{quicktipbox
To find the mean of a frequency distribution, use the formula \( Mean = \frac{\sum (f \cdot M)}{\sum f} \), where \( f \) is the frequency and \( M \) is the mid-point of each class interval.
\end{quicktipbox Quick Tip: To find the mean of a frequency distribution, use the formula \( Mean = \frac{\sum (f \cdot M)}{\sum f} \), where \( f \) is the frequency and \( M \) is the mid-point of each class interval.

We are given:
- Slant height \( l = 4 \, cm \),
- Perimeters (circumferences) of the circular ends \( 18 \, cm \) and \( 6 \, cm \).

Step 1: Find the radii of the circular ends.
The perimeter (circumference) of a circle is given by \( C = 2\pi r \), where \( r \) is the radius.

For the larger circle, \( C_1 = 18 \, cm \): \[ 18 = 2\pi r_1 \quad \Rightarrow \quad r_1 = \frac{18}{2\pi} = \frac{9}{\pi} \approx 2.87 \, cm. \]

For the smaller circle, \( C_2 = 6 \, cm \): \[ 6 = 2\pi r_2 \quad \Rightarrow \quad r_2 = \frac{6}{2\pi} = \frac{3}{\pi} \approx 0.95 \, cm. \]

Step 2: Use the formula for the curved surface area of the frustum.
The formula for the curved surface area of a frustum of a cone is: \[ A = \pi (r_1 + r_2) l, \]
where \( r_1 \) and \( r_2 \) are the radii of the circular ends, and \( l \) is the slant height.

Substitute the values: \[ A = \pi (2.87 + 0.95) \times 4 \approx \pi \times 3.82 \times 4 = 15.28\pi \approx 47.98 \, cm^2. \]


Final Answer:
The curved surface area of the frustum is \( \boxed{47.98 \, cm^2} \).


% Quicktip
\begin{quicktipbox
The curved surface area of a frustum of a cone is given by \( A = \pi (r_1 + r_2) l \), where \( r_1 \) and \( r_2 \) are the radii of the circular ends and \( l \) is the slant height.
\end{quicktipbox Quick Tip: The curved surface area of a frustum of a cone is given by \( A = \pi (r_1 + r_2) l \), where \( r_1 \) and \( r_2 \) are the radii of the circular ends and \( l \) is the slant height.