Question

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

Answer 1

Given,

$\sin 3\theta =\cos (\theta -6^o)$

$\cos ({90}^o-3\theta )=\cos (\theta -6^o)$ ...........$\because \cos (90-x)=\sin x$

${90}^o-3\theta =\theta -6^o$

$4\theta ={96}^o$

$\theta ={24}^o$