Question

If A and B are acute angles such that 0°< A+B < 90° and sin(A + B) = 1 and cos (A-B)=√3/2 Then find angles A and B.​

Answer 1

Answer:

A = 60°,B = 30°.

Step-by-step explanation:

Given,

sin(A+B) = 1

sin(A+B) = sin 90°

(Since both sides 'sin' cancel)

A+B = 90° → (1)

Also given,

cos(A-B) = √3/2

cos(A-B) = cos 30°

(Since both sides 'cos' cancel)

A-B = 30° → (2)

By solving equation (1)&(2)

A + B = 90°

A - B = 30°

________

2A = 120° (Since +B&-B are cancel)

A = 120°/2

:. A = 60°

Substitute 'A' value in equation (1)

A + B = 90°

=> 60° + B = 90°

=> B = 90° - 60°

:. B = 30°

:. A = 60°,B = 30°.

I hope it helps you.