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.

    In the adjoining figure, $\triangle CAB$ is a right triangle, right angled at A and $AD \perp BC$. Prove that $\triangle ADB \sim \triangle CDA$. Further, if $BC = 10$ cm and $CD = 2$ cm, find the length of AD.


      • 2.
        The perimeters of two similar triangles are 22 cm and 33 cm respectively. If one side of the first triangle is 9 cm, then find the length of the corresponding side of the second triangle.


          • 3.
            In a trapezium \(ABCD\), \(AB \parallel DC\) and its diagonals intersect at \(O\). Prove that \[ \frac{OA}{OC} = \frac{OB}{OD} \]


              • 4.
                Three coins are tossed together. The probability that at least one head comes up is

                  • \(\dfrac{3}{8}\)
                  • \(\dfrac{7}{8}\)
                  • \(\dfrac{1}{8}\)
                  • \(\dfrac{3}{4}\)

                • 5.
                  Find the sum of first 20 terms of an A.P. whose n\(^{th}\) term is given by \(a_n = 5 + 2n\). Can 52 be a term of this A.P. ?


                    • 6.
                      Using prime factorisation, find the HCF of 144, 180 and 192.

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