NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.2

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NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.2 covers different cases of the relationship between Zeroes and Coefficients of a Polynomial. The exercise has 2 questions with 6 cases each.

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Check out other exercise solutions of Class 10 Maths Chapter 2 Polynomials

Class 10 Chapter 2 Polynomials Topics:

Degree of Polynomial	Zeroes of a Polynomial	Polynomials Formula
Vietas Formula	Roots of Polynomials	Cubic Polynomial
Polynomials Revision Notes	Polynomials Important Questions	Parabola Graph

CBSE Class 10 Maths Study Guides:

Math Study Notes	Difference between in Maths	Math MCQs
Calculus	Math Formula	Mensuration
NCERT Solutions for Class 10 Maths	Arithmetic	Trigonometry

CBSE X Related Questions

1.
A kite is flying at a height of $60 \text{ m}$ above the ground level. Ravi, standing at the roof of the house is holding the string straight and observes the angle of elevation of kite as $30^{\circ}$. From the bottom of the same building, the angle of elevation of kite is $45^{\circ}$. Find the length of the string and height of roof from the ground. (Use $\sqrt{3} = 1.73$)
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2.
Use graphical method to solve the system of linear equations : $x = -3$ and $5x - 2y = -5$.
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3.
PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If OP = 13 cm, then find the length AB and PA.
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4.
Prove that :
$\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta$.
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5.
$ABCD$ is a parallelogram such that $AF = 7 \text{ cm}$, $FB = 3 \text{ cm}$ and $EF = 4 \text{ cm}$, length $FD$ equals
- $\frac{21}{4} \text{ cm}$
- $\frac{28}{3} \text{ cm}$
- $\frac{12}{7} \text{ cm}$
- $5.5 \text{ cm}$
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6.
Prove that: $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta$
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NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.2

CBSE X Related Questions

2.
Use graphical method to solve the system of linear equations : $x = -3$ and $5x - 2y = -5$.

3.
PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If OP = 13 cm, then find the length AB and PA.

4.
Prove that :
\(\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta\).

5.
\(ABCD\) is a parallelogram such that \(AF = 7 \text{ cm}\), \(FB = 3 \text{ cm}\) and \(EF = 4 \text{ cm}\), length \(FD\) equals

6.
Prove that: $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta$

Similar Mathematics Concepts

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NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.2

CBSE X Related Questions

2.Use graphical method to solve the system of linear equations : $x = -3$ and $5x - 2y = -5$.

3.PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If OP = 13 cm, then find the length AB and PA.

4.Prove that : \(\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta\).

5.\(ABCD\) is a parallelogram such that \(AF = 7 \text{ cm}\), \(FB = 3 \text{ cm}\) and \(EF = 4 \text{ cm}\), length \(FD\) equals

6.Prove that: $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta$

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

2.
Use graphical method to solve the system of linear equations : $x = -3$ and $5x - 2y = -5$.

3.
PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If OP = 13 cm, then find the length AB and PA.

4.
Prove that :
\(\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta\).

5.
\(ABCD\) is a parallelogram such that \(AF = 7 \text{ cm}\), \(FB = 3 \text{ cm}\) and \(EF = 4 \text{ cm}\), length \(FD\) equals

6.
Prove that: $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta$