Question

In a triangle ABC, sin ( A+2B ) = 1 and cos (A- B) = 1 then find A+B.​

Answer 1

A = 67.5°

B = 37.5°

C = 75°

Given:

$\sin (A+B-C)=\frac{1}{2}$

$\cos (B+C-A)=\frac{1}{\sqrt{2}}$

Step-by-step explanation:

We know that in a triangle, sum of the angles = 180°

A+B+C = 180 → (1)

We know that,

$\sin 30=\frac{1}{2}$

$\cos 45=\frac{1}{\sqrt{2}}$

So,

sin (A+B-C) = sin 30

A+B-C = 30 → (2)

And

cos (B+C-A) = cos 45

B+C-A = 45 → (3)

On solving equation (1) and (2), we get,  

A+B+C-A-B+C = 180-30 = 150

2C = 150

C = 75°

Substituting C=75 in equation (2), we get,

A+B-75 = 30

A+B = 105 → (4)

Also, substituting in equation (3), we get,

B+75-A =45

A-B = 30 → (5)

Adding equations (4) and (5), we get,

2A = 135 → A = 67.5°

B = A-30 = 67.5 - 30 = 37.5°

Therefore, A=67.5°; B=37.5°; and C=75°