Question

If $\sin 3\theta =\cos (\theta -6^{\circ })$ and $\theta -6^{\circ }$ are acute angles,find the value of $\theta$

Answer 1

We have $\sin 3\theta =\cos (\theta -6^{\circ })$

and $\cos ({90}^{\circ }-3\theta )=\cos (\theta -6^{\circ })$

$\Rightarrow {90}^{\circ }-3\theta =\theta -6^{\circ }$

$\Rightarrow 4\theta ={96}^{\circ }$

$\therefore \theta =\frac{{96}^{\circ }}{4}={24}^{\circ }$