Question

If $\cos 7A=\sin (A-6^o)$, where $7A$ is an acute angle. Find the value of $A$.

Answer 1

When angle is acute,

$\cos \theta =\sin ({90}^{\circ }-\theta )$

So $\cos 7A=\sin ({90}^{\circ }-7A)$

$\Rightarrow \sin ({90}^{\circ }-7A)=\sin (A-6^{\circ })$

$\Rightarrow {90}^{\circ }-7A=A-6^{\circ }$

$\Rightarrow 8A={96}^{\circ }\Rightarrow A={12}^{\circ }$