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

Given: sin3A = cos(A – 10°)

$\sin 3 A=\cos \left(A-10^{\circ}\right)$

$\Rightarrow \cos \left(90^{\circ}-3 A\right)=\cos \left(A-10^{\circ}\right) \quad\left(\because \sin \theta=\cos \left(90^{\circ}-\theta\right)\right)$

$\Rightarrow 90^{\circ}-3 A=A-10^{\circ}$

$\Rightarrow 90^{\circ}+10^{\circ}=A+3 A$

$\Rightarrow 4 A=100^{\circ}$

$\Rightarrow A=\frac{100^{\circ}}{4}$

$\Rightarrow A=25^{\circ}$

Hence, $\angle A=25^{\circ}$.