Question

If $\sin 3A=\cos (A-{26}^o)$, where $3A$ is an acute angle, find the value of $A$.

Answer 1

Given $sin3A=cos(A-{26}^{\circ })$

$\Rightarrow cos({90}^{\circ }-3A)=cos(A-{26}^{\circ })$

Equating

${90}^{\circ }-3A=A-{26}^{\circ }$

$\Rightarrow {90}^{\circ }+{26}^{\circ }=4A$

$\Rightarrow 4A={116}^{\circ }$

$\Rightarrow A=\frac{{116}^{\circ }}{4}$

$\Rightarrow A={29}^{\circ }$