Question

If $\sin 3A=\cos (A-{10}^{\circ })$, where $3A$ is an acute angle, then $\angle A=......$

• A. ${35}^{\circ }$
• B. ${25}^{\circ }$
• C. ${20}^{\circ }$
• D. ${45}^{\circ }$

Answer 1

The correct option is B ${25}^{\circ }$

$\sin 3A=\cos (A-{10}^o)\Rightarrow \cos ({90}^o-3A)=\cos (A-{10}^o)\Rightarrow {90}^o+{10}^o=3A+A\Rightarrow {100}^o=4A\Rightarrow A={25}^o$

Therefore, Answer is ${25}^o$