If 2θ + 45° and 30° − θ are acute angles,

Given that: $\sin \left(2 \theta+45^{\circ}\right)=\cos \left(30^{\circ}-\theta\right)$ where $\left(2 \theta+45^{\circ}\right)$ and $\left(30^{\circ}-\theta\right)$ are acute angles

We have to find $\theta$

So we have

$\sin \left(2 \theta+45^{\circ}\right)=\cos \left(30^{\circ}-\theta\right)$

$\Rightarrow \sin \left(2 \theta+45^{\circ}\right)=\sin \left[90^{\circ}-\left(30^{\circ}-\theta\right)\right]$

$\Rightarrow 2 \theta+45^{\circ}=90^{\circ}-30^{\circ}+\theta$

$\Rightarrow \theta=15^{\circ}$

Hence the value of $\theta$ is $\theta=15^{\circ}$