If cos 2A = sin (A – 15°), where 2A is acute then find ∠A.

Given: cos2A = sin(A – 15°)

$\cos 2 A=\sin \left(A-15^{\circ}\right)$

$\Rightarrow \sin \left(90^{\circ}-2 A\right)=\sin \left(A-15^{\circ}\right) \quad\left(\because \cos \theta=\sin \left(90^{\circ}-\theta\right)\right)$

$\Rightarrow 90^{\circ}-2 A=A-15^{\circ}$

$\Rightarrow 90^{\circ}+15^{\circ}=A+2 A$

$\Rightarrow 3 A=105^{\circ}$

$\Rightarrow A=\frac{105^{\circ}}{3}$

$\Rightarrow A=35^{\circ}$

Hence, $\angle A=35^{\circ}$.