Question

An acute angle ABC. If sin(A+B-C) 1/2 cos (B+C-A). Find A, B & C​

Answer 1

Answer:

We have sin (A + B+ C) = 1/2 = sin 45° Read more on Sarthaks.com - https://www.sarthaks.com/176851/in-an-acute-angled-triangle-abc-if-sin-a-b-c-1-2-and-cos-b-c-a-1-2-find-a-b-and-c

Answer 2

Answer:

Angles of ‘A’, ‘B’ and ‘C’ are 67.5 , 37.5 and 75 respectively.

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 = 1/2

cos 45 = 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°