Sin 30 Degrees: Value, Derivation & Trigonometry Table

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Sin 30 degrees has a value of ½ or 0.5. In radians, Sin 30is represented as Sin \(\frac{\pi}{6}\). Sine or Sin is one of the most important trigonometric ratios whose values range between -1 to +1. The other important trigonometric ratios are Cos, Tan, Sec, Cosec, Cot. In this article, we will learn about the value of Sin 30and its derivation.

Sin 30o = ½ 

Read Here: Applications of Trigonometry

Key Takeaways: Sin 30 degrees, Radians, Trigonometric Ratios, Trigonometric Table, Geometric Derivation.


Sin 30 Degrees Value 

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In a right-angled triangle, the sin of the angle is the ratio of the side opposite the angle to the hypotenuse (longest side of triangle).

\(Sin \theta = \frac{Opposite}{Hypotenuse}\)

The angle sin 30° is between 0° and 90° and lies in the First Quadrant and the value for sin 30 degree angle is 0.5.

Sin 30 degrees in first quadrant

Sin 30 Degrees

Read Here: Introduction to Trigonometry


Geometric Derivation of Sin 30 Degree 

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Consider an equilateral triangle ABC

We know that, All angles in an equilateral triangle are 60 degrees.

Thus , ∠A = ∠B = ∠C = 60o

Now draw a perpendicular line AD from A to BC as shown in the figure below.

Geometric Derivation of Sin 30 Degree
Geometric Derivation of Sin 30 Degree

Now in Triangle ΔABD = ΔACD

Therefore,       BD = DC

Also ΔBAD = ΔCAD (By Corresponding Parts of Congruent Triangles)

Lets assume the length of side of the equilateral triangle  AB = BC = AC = 2a.

Then, BD = ½ BC = ½ x 2a = a

 In Right Triangle ADB,

AD2 + BD2 = AB2 [By Pythagoras theorem]

 AD2 = = AB2- BD2

(2a)2 – a2 = (3a)2

AD2 = 3a2

AD = a√3

Now, In right Triangle ADB,

Sin 30o = BD/AB = a/2a = ½ or 0.5.

Similarly, we can represent Sin 30 by using trigonometric identities as follows.

  • Sin (180– 30o) = Sin 50o
  • -Sin (180o + 30o) = -Sin 210o
  • Cos (90o - 30o) = Cos 60o
  • -Cos (90o + 30o) = -Cos 120o

Check More: Trigonometry Important Formulae


Trigonometric Values Table 

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The important values of trigonometric ratios are given in the table below.

Angles 

0o

0

30o

\(\frac{\pi}{6}\)

45o

\(\frac{\pi}{4}\)

60o

\(\frac{\pi}{3}\)

90o

\(\frac{\pi}{2}\)

180o

\(\pi\)

Sin 0 ½  \(\frac{1}{\sqrt{2}}\) \(\frac{\sqrt{3}}{2}\) 1 0
Cos 1 \(\frac{\sqrt{3}}{2}\) \(\frac{1}{\sqrt{2}}\) ½  0 -1
Tan 0 \(\frac{1}{\sqrt{3}}\) 1 \(\sqrt{3}\) Not Defined 0
Cosec Not Defined 2 \(\sqrt{2}\) \(\frac{2}{\sqrt{3}}\) 1 Not Defined
Sec 1 \(\frac{2}{\sqrt{3}}\) \(\sqrt{2}\) 2 Not Defined -1
Cot Not Defined \(\sqrt{3}\) 1 \(\frac{1}{\sqrt{3}}\) 0 Not Defined

Also Read:


Things to Remember 

  • The value of Sin 30 degrees is ½ or 0.5.
  • Sin 30 degrees is also written as Sin \(\frac{\pi}{6}\) or sin (0.523598..)
  • The decimal value of sin 30 degree and Cos 60 degree is same i.e. 0.50
  • Sin 30° is lying in the first quadrant so its final value will be positive.
  • Sinθ reciprocal is Cosecθ. We also say that Sinθ = Perpendicular / Hypotenuse, then Cosecθ = Hypotenuse / Perpendicular.
  • For the conversion of degree to radian we use, θ in radians = θ in degree* (Pi/180).

Read More: Cosine Rule Important Notes


Sample Questions

Ques: In a Right angle triangle ABC, Side AB = 24cm & side BC = 7 cm, Find 
1) Sin A & Cos A
2) Cos C & Sin C (2015 CBSE, 3 marks)
AB = 24cm & side BC = 7 cm

Ans: In triangle ABC,

Given: side AB = 24cm & BC = 7 cm

By using Pythagoras theorem,

AC2 = AB2+ BC2

AC2 = 242+ 72

AC2 = 576 + 49 = 625

AC = 25

Now for,

 1). Sin A = BC/AC = 7/25

Cos A = AB/AC = 24/25.

2). Cos C =BC/AC = 7/25

Sin C = AB/AC =24/25.

Ques: In a Right angle triangle ABC, Side AB = 30cm & side BC = 10 cm, Find Sin A & Cos A (2 marks)

Ans: In triangle ABC,

Given: side AB = 30 cm & BC = 10 cm

By using Pythagoras theorem,

AC2 = AB2+ BC2

AC2 = 302+ 102

AC2 = 900 + 100 = 1000

AC = 31.62

Now for,

Sin A = BC/AC = 10/31.62 = 0.316

Cos A = AB/AC = 30/31.62 = 0.94

Ques: Find the value of Sin230 + tan245+ cot2 60 (2 marks)

Ans: From the trigonometric table, 

We know that, 

Sin 30 = ½

Tan 45 = 1

Cot 60 = \(\frac{1}{\sqrt{3}}\)

Therefore,

→ (1/2)2 + 12 +( \(\frac{1}{\sqrt{3}}\))2

→ ¼ + 1 + 1/3 

= 1.58

Ques: Find the value of Sin230 + Sin245+ Sin2 60 (2 marks)

Ans: From the trigonometric table, 

We know that, 

Sin 30 = ½

Sin 45 = \(\frac{1}{\sqrt{2}}\)

Sin 60 = \(\frac{\sqrt{3}}{2}\)

Therefore,

→ (1/2)2 + (\(\frac{1}{\sqrt{2}}\))2 +(\(\frac{\sqrt{3}}{2}\))2

→ ¼ + ½  + ¾  

= 1.5

Ques: Find tan A – cot C in figure? (3 marks)
tan A – cot C

Ans: In right angle triangle, ABC

By using Pythagoras theorem, we get

BC2 = AC2 – AB2

QR2 = 132 – 122

 = (13-12) (13+12)

 = 1* 25 = 25

BC2 = 25

BC = 5

Now,

Tan A = BC/AB = 5/12

Cot C = BC/AB = 5/12

So, tan A – cot C = 5/12 - 5/12 = 0

Ques: What is the value of sin(-30)? [CBSE, 2014, 2 marks]

Ans: We already know that the value of sin 30 = 0.5

 sin (-30) = - sin (30)

therefore, sin (-30) = - 0.5

Ques: What is the value of sin 30 + cos 30? (2 marks)

Ans: We already know from the trigonometric table,

Sin 30 = ½ 

Cos 30 = \(\frac{\sqrt{3}}{2}\)

Therefore,

½ + \(\frac{\sqrt{3}}{2}\) = 1.36

Ques: What is the value of sin 30 + tan 30? (2 marks)

Ans: We already know from the trigonometric table,

Sin 30 = ½ 

Tan 30 = \(\frac{1}{\sqrt{3}}\)

Therefore,

½ + \(\frac{1}{\sqrt{3}}\) = 1.07

Ques: What is the value of sin 30 + Cos 60? (2 marks)

Ans: We already know from the trigonometric table,

Sin 30 = ½ 

Cos 60 = ½ 

Therefore,

½ + ½  = ¼ 

= 0.25

Ques: In right angled triangle ABC, Angle ACB = 30° and AB = 5 cm. Find the lengths of AC and BC sides? [CBSE, 2018, 3 marks]
Angle ACB = 30° and AB = 5 cm

Ans: Given: ∠ACB = 30° and side AB = 5 cm.

We know that,

Tan θ = Perpendicular/ Base

So, Tan C = AB/ BC or tan 30° = 5/ BC

1/ √3 = 5cm /BC (by using table, tan30° = 1/√3)

And also, BC = 5√3 cm.

Now, for third side by using trigonometric identities we get

Sin C = AB/ AC or Sin 30° = 5cm / AC

We know that Sin 30° = ½.

So, ½ = 5 cm /AC

AC = 10 cm.


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CBSE X Related Questions

  • 1.

    On the day of her examination, Riya sharpened her pencil from both ends as shown below.

    The diameter of the cylindrical and conical part of the pencil is 4.2 mm. If the height of each conical part is 2.8 mm and the length of the entire pencil is 105.6 mm, find the total surface area of the pencil.


      • 2.

        Find the mean and mode of the following data:

        Class15--2020--2525--3030--3535--4040--45
        Frequency1210151175


          • 3.
            A peacock sitting on the top of a tree of height 10 m observes a snake moving on the ground. If the snake is $10\sqrt{3}$ m away from the base of the tree, then angle of depression of the snake from the eye of the peacock is

              • $60^\circ$
                 

              • $45^\circ$
              • $30^\circ$
              • $90^\circ$

            • 4.
              The given figure shows a circle with centre O and radius 4 cm circumscribed by \(\triangle ABC\). BC touches the circle at D such that BD = 6 cm, DC = 10 cm. Find the length of AE.
               BC touches the circle at D such that BD = 6 cm


                • 5.

                  In the adjoining figure, TP and TQ are tangents drawn to a circle with centre O. If $\angle OPQ = 15^\circ$ and $\angle PTQ = \theta$, then find the value of $\sin 2\theta$.


                    • 6.

                      Given that $\sin \theta + \cos \theta = x$, prove that $\sin^4 \theta + \cos^4 \theta = \dfrac{2 - (x^2 - 1)^2}{2}$.

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