Question

If sin (A + B) = 1 and `tan (A - B) = 1/sqrt(3)`, 0° < (A + B) ≤ 90° and A > B then find the values of A and B.

Answer 1

Given:

sin (A + B) = 1

`tan(A - B) = 1/sqrt(3)`

0° < (A + B) ≤ 90° and A > B

Step-wise calculation:

1. sin (A + B) = 1 and 0° < (A + B) ≤ 90° 

⇒ A + B = 90°

2. `tan (A - B) = 1/sqrt(3)` 

⇒ A – B = 30°   ...(Since `tan 30^circ = 1/sqrt(3)` and A > B so A – B is positive)

3. Solve the linear system:

A + B = 90°

A – B = 30°

Add: 2A = 120°

⇒ A = 60°

Subtract: 2B = 60°

⇒ B = 30°