Question

If `sin (A + B) = sqrt(3)/2` and `cos (A - B) = sqrt(3)/2`, 0° < (A + B) ≤ 90° and A > B then find the values of A and B.

Answer 1

Given: `sin(A + B) = sqrt(3)/2, cos(A - B) = sqrt(3)/2`, 0° < (A + B) ≤ 90° and A > B.

Step-wise calculation:

1. Since 0° < A + B ≤ 90° and `sin (A + B) = sqrt(3)/2`, we take A + B = 60°   ...`(∵ sin 60^circ = sqrt(3)/2)`

2. `cos (A - B) = sqrt(3)/2` 

⇒ A – B = 30°   ...`(cos 30^circ = sqrt(3)/2)` 

Because A > B, A – B is positive, so we take the +30° solution.

3. Solve the linear system:

A + B = 60°

A – B = 30°

Add: 2A = 90°

⇒ A = 45° 

Then B = 60° – A

= 60° – 45°

= 15°