Question

Q9. sin (A + B) = 1 and sin (A – B) =1/2 ; 0° < A + B ≤ 90°; ∠A > ∠B. Find ∠A and ∠B

Answer 1

EXPLANATION.

=> Sin ( A + B) = 1

=> Sin ( A - B) = ½

=> 0° < A + B ≤ 90° and A > B

To find A and B.

=> Sin ( A + B) = sin 90° ......(1)

=> Sin ( A - B) = Sin 30° .......(2)

From equation (1) and (2) we get,

=> A + B = 90°

=> A - B = 30°

=> 2A = 120°

=> A = 60°

put the value of A = 60° in equation (1)

we get,

=> 60° + B = 90°

=> B = 30°

Therefore,

=> A = 60° and B = 30°

Answer 2

Correct Question:

If sin(A + B) = 1 and sin (A – B) = 1/2, 0 ≤ A + B ≤ 90° and A > B, then find A and B.

Given information:

sin(A + B) = 1

sin(A – B) = 1/2

0 ≤ A + B ≤ 90°

To find:

Value of A and B

Solution:

sin(A + B) = 1 = sin 90°

A + B = 90° ..........(i)

sin(A – B) = 1/2 = sin 30°

A – B = 30° ........(ii)

Solving equation (i) and (ii), we get

A = 60° and B = 30°