Question

If sin 3A = cos (A – 10°), where 3A is an acute angle then find ∠A.

Answer 1

Given: sin3A = cos(A – 10°)

$$\begin{gathered} \sin 3A=\cos \left(A-10°\right) \\ \Rightarrow \cos \left(90°-3A\right)=\cos \left(A-10°\right)\left(\because \sin \theta =\cos \left(90°-\theta \right)\right) \\ \Rightarrow 90°-3A=A-10° \\ \Rightarrow 90°+10°=A+3A \\ \Rightarrow 4A=100° \\ \Rightarrow A=\frac{100°}{4} \\ \Rightarrow A=25° \\ \mathrm{Hence},\angle A=25°. \end{gathered}$$