Law of Sines

To find the angles of a triangle, we use the Law of sines. The trigonometry ratios such as sine, cosine and tangent are primary functions which are used to find the unknown angles or sides of a triangle. The applications of sine law are given below:

  • It can be used to compute the other sides of a triangle when two angles and one side is given.
  • When two sides and one non-included angle is given.

Law of sines can be used for all types of triangles such as an acute, obtuse and right triangle. Let us learn here how to prove this law.


In general, the law of sines is defined as the ratio of side length to the sine of the opposite angle. It holds for all the three sides of a triangle respective of their sides and angles.

\(\frac{a}{Sin A}=\frac{b}{Sin B}=\frac{c}{Sin C}\)

Law of Sines

So, we use the Sine rule to find unknown lengths or angles of the triangle. It is also called as Sine Rule, Sine Law or Sine Formula. 

Also, read:

Law of Sines Formula

The formulas used with respect to law of sine are given below.

a / Sin A= B / Sin B= c / Sin C
a: b: c = Sin A: Sin B: Sin C
a / b = Sin A / Sin B
b / c = Sin B / Sin C

It denotes that if we divide side a by the Sine of A, it is equal to the division of side b by the Sine of B and also equal to the division of side c by Sine of C (Or) The sides of a triangle are to one another in the same ratio as the sines of their opposite angles.

Here, Sin A is a number and a is the length.

Law of Sine Proof

We need a right-angled triangle to prove the above as the trigonometric functions are mostly defined in terms of this type of triangle only.

Given: ABC.

Construction: Draw a perpendicular, CD AB. Then CD = h is the height of the triangle. “h” separates the ABC in two right-angled triangles, CDA and CDB.

To Show: a / b = Sin A / Sin B

Proof: In the  CDA,

           Sin A= h/b

           And in  CDB,

           Sin B = h/a

Therefore, Sin A / Sin B = h / a / h / b= h / b × a / h= a/b

And we proved it.

Similarly, we can prove, Sin B/ Sin C= b / c and so on for any pair of angles and their opposite sides.

Example of Sine Law

Question: Solve △PQR in which ∠ P = 63.5° and ∠Q = 51.2°  and r = 6.3 cm.

Law of Sines-Example


Let’s Calculate the third angle-

∠R = 180° – 63.5°– 51.2 °= 65.3°

Now, Let’s calculate the sides:

6.3 / Sin 65.3 = p / Sin 63.5 (BC= p)

p = (6.3 × Sin 63.5) / Sin 65.3

p= 6.21 cm approximately.

Similarly, 6.3 / Sin 65.3 = q / Sin 51.2 (q= AB)

q = (6.3 × Sin51.2) / Sin 65.3

q= 5.40 cm

∠R= 65.3°

p=6.21 cm

q=5.40 cm

Get started with Practicing more of Trigonometric problems with BYJU’S.

Leave a Comment

Your email address will not be published. Required fields are marked *