If $2 \theta+45^{\circ}$ and $30^{\circ}-\theta$ are acute angles, find the degree measure of $\theta$ satisfying $\sin \left(2 \theta+45^{\circ}\right)=\cos \left(30^{\circ}-\theta\right)$.
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 $\theta=15^{\circ}$
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