Question

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

Answer 1

You should know that , $cos\left(x\right)=sin(90-x)$

hence $cos(A-26)=sin(90-A+26)$

$=sin(116-A)$

Now, since $3A$ is acute,

Hence $A$ is also acute and hence we can directly remove sin function.

$sin3A=sin(116-A)$

$3A=116-A$

$4A=116$

Hence $A=29$