Question

If sin 3A = cos (A − 10°), where 3A is an acute angle, then ∠A = ?
(a) 35°
(b) 25°
(c) 20°
(d) 45°

Answer 1

$$\begin{gathered} \text{(b) 25°} \\ \text{We have:} \\ \left[\sin 3\text{A=cos}(\text{A}-{10}^0)\right] \\ =>\cos ({90}^0-3\text{A})=\cos (\text{A}-{10}^0)[\because \sin \theta =\cos ({90}^0-\theta )] \\ =>{90}^0-3\text{A}=\text{A}-{10}^0 \\ =>-4\text{A=}-100 \\ =>\text{A=}\frac{\sout{1}\sout{0}{\sout{0}}^{25}}{{\sout{4}}_1} \\ =>{\text{A=25}}^0 \end{gathered}$$