1.

In an acute angle triangle ABC, sin (A + B - C) = 12, cot (A - B + C) = 0 and cos (B + C – A) =12. What are the values of A, B, and C?​

Answer»

Step-by-step explanation:

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

A+B+C = 180 → (1)

sin 30

1 \div 2 \\

cos45

1 \div \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°


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