Question

$\sin 4A=\cos (A-{20}^o)$, where 4A is an acute angle, find the value of A in degrees

Answer 1

We know that,

$\sin A=\cos ({90}^o-A)$

Now, $\sin 4A=cos(A-{20}^o)\Rightarrow \cos ({90}^o-4A)=cos(A-{20}^o)\Rightarrow {90}^o-4A=A-{20}^o\Rightarrow {110}^o=5A\Rightarrow A={22}^o$