Question

If $sin3A=cos(A-{10}^o)$ and $3A$ is acute then, find $\angle A$ .

• A. ${25}^o$
• B. ${30}^o$
• C. ${50}^o$
• D. ${15}^o$

Answer 1

The correct option is A ${25}^o$

Given: $sin3A=cos(A-{10}^o)$

$sin3A=cos({90}^o-3A)....(\because sin\theta =cos({90}^o-\theta \left)\right)$

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