Question

If $\sin \theta =\cos (\theta -6^o)$, find the value of $\theta$.

Answer 1

We have,

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

$\Rightarrow \cos ({90}^o-\theta )=\cos (\theta -6^o)$

Now, Comparing that,

$({90}^o-\theta )=(\theta -6^o)$

$\Rightarrow {90}^o+6^o=2\theta$

$\Rightarrow 2\theta ={96}^o$

$\Rightarrow \theta ={48}^o$

Hence, this is the answer.