Question

Find acute angles A and B, if sin (A + 2B) =√3/2 and cos (A + 4B) = 0, A > B.

Answer 1

A = 30° ; B = 15°

Step-by-step explanation:

Given: sin (A + 2B) =√3/2 and cos (A + 4B) = 0, A > B. A & B are acute angles.

Find: A and B

Solution:

 sin (A + 2B) = √3/2

 Sin 60° = √3/2

 So we get A + 2B = 60° -------(1)

 cos (A + 4B) = 10

 cos 90° = 0

 So we get A + 4B = 90° --------(2)

 Subtracing (1) from (2), we get: 2B = 30°

 Therefore B = 15°.

 Substituting B in (1), we get A + 30° = 60°  

 Therefore A = 30°