Question

in  an acute angled triangle ABC,if cos(A+b-c) = 1/root2 and sin(b+c-a)=root3/2,then find the value of a,b and c

Answer 1

Given,

Triangle is acute angle angled triangle.

Also given that,

cos(a+b-c) = \(\frac{1}{\sqrt2}\) = cos 45°

And,

sin(b+c-a) = \(\frac{\sqrt3}{2}\) = sin 60°

∴ a+b-c = 45° ....(1)

And,

b+c-a = 60° ....(2)

By adding (1) & (2) we get,

2b = 105°

⇒ b = \(\frac{105°}{2}\)= 52.5°

Also,

a+b+c = 180° ...(3)

(∵ sums of angles is 180°)

By (3) - (1) we get,

2c = 135°

⇒ c = \(\frac{135°}{2}\) = 67.5°

By (3) - (2) we get,

2a = 120°

⇒ a = 60°

Hence,

a = 60°, b = 52.5° and c = 67.5°