Question

Find acute angles A & B, if sin (A + 2B) = ${\displaystyle \frac{\sqrt{3}}{2}\cos (A+4B)=0,A>B}$

Answer 1

Sin (A + 2B) = Sin 60°

Cos (A + 4B) = cos 90°

A + 2B = 60° …(i)

A + 4B = 90° …(ii)

Subtracting (ii) from (i)

A + 4B = 90°

${\displaystyle \frac{-A–2B=-60}{2B=30°}}$  ∴ B = 15°

A + 4B = 90°

4B = 4(15°) = 4B = 60°

∴ A + 60° = 90° ∴ A = 30°