Question

If sin(A + B) = 1 and cos(A – B)= `sqrt(3)/2`, 0° < A + B ≤ 90° and A > B, then find the measures of angles A and B.

Answer 1

	0°	30°	45°	60°	90°
sin	0	`1/2`	`1/sqrt2`	`sqrt3/2`	1
cos	1	`sqrt3/2`	`1/sqrt2`	`1/2`	0
tan	0	`1/sqrt3`	1	`sqrt3`	Not def.

Given that

sin(A + B) = 1

But we know that

sin 90° = 1

Thus, sin (A + B) = sin 90°

∴ A + B = 90°  ......(1)

cos(A – B)= `sqrt(3)/2`

But we know that

cos 30° = `sqrt(3)/2`

Thus, cos(A – B) = 30° 

∴ A – B = 30°  ......(2)

Our equations are

A + B = 90°  ......(1)

A – B = 30°  ......(2)

Adding (1) and (2)

A + B + A  – B = 90° + 30°

2A = 120°

A = `120^circ/2`

A = 60°

Putting A = 60° in (1)

A + B = 90°

60° + B = 90°

B = 90° – 60°

B = 30°

Hence A = 60°, B = 30°