Question

In any triangle ABC, prove that :

sinA/a = sinB + sinC / b+c = sinB-sinC / b-c​

Answer 1

In any ∆ABC, using sine rule we say:

a/sinA = b/sinB = c/sinC, where a, b, c are the sides opposite to angle A, B, and C respectively.

Let all these be equal to k.

a/sinA = b/sinB = c/sinC = k

=> a = ksinA ; b = ksinB ; c = ksinC

Then,

• a = ksinA

=> 1/k = sinA/a ... (1)

• b + c = ksinB + ksinC

=> b + c = k(sinB + sinC)

=> 1/k = (sinB + sinC)/(b + c) ...(2)

• b - c = ksinB - ksinC

=> b - c = k(sinB - sinC)

=> 1/k = (sinB - sinC)/(b - c) ...(3)

Therefore, from (1), (2) & (3):

sinA/a=(sinB+sinC)/(b+c)=(sinB-sinC)/(b-c)

Proved.

Answer 2

In any triangle ABC,

sinA/a = sinB + sinC / b+c = sinB-sinC / b-c