Question

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

Answer 1

Given that

$\sin 3A=\cos (A-26)$

$\cos (90-3A)=\cos (A-26)$ $[\sin \theta =\cos (90-\theta \left)\right]$

Comparing angles

$90-3A=A-26$

$-3A-A=-26-90$

$-4A=-116$

$A=\frac{116}{4}$

$A={29}^o$