If sin (θ + 36°) = cos θ, where (θ + 36°) is acute, find θ.

Given: sin(θ + 36°) = cosθ

$\sin \left(\theta+36^{\circ}\right)=\cos \theta$

$\Rightarrow \cos \left(90^{\circ}-\left(\theta+36^{\circ}\right)\right)=\cos \theta \quad\left(\because \sin \theta=\cos \left(90^{\circ}-\theta\right)\right)$

$\Rightarrow \cos \left(90^{\circ}-\theta-36^{\circ}\right)=\cos \theta$

$\Rightarrow 54^{\circ}-\theta=\theta$

$\Rightarrow \theta+\theta=54^{\circ}$

$\Rightarrow 2 \theta=54^{\circ}$

$\Rightarrow \theta=\frac{54^{\circ}}{2}$

$\Rightarrow \theta=27^{\circ}$

Hence, $\theta=27^{\circ}$.