Question

if $\alpha$ and $\beta$ are angles in the first quadrant, $\tan \alpha =\frac{1}{7},\sin \beta =\frac{1}{\sqrt{10}}$, then using the formula $\sin (A+B)=\sin A\cos B+\cos A\sin B,$ one can find the value of $(\alpha +2\beta )$ to be

• A. 0${}^{\circ }$
• B. 45${}^{\circ }$
• C. 60${}^{\circ }$
• D. 90${}^{\circ }$

Answer 1

The correct option is B 45${}^{\circ }$

Given

$\tan \alpha =\frac{1}{7},\sin \beta =\frac{1}{\sqrt{10}}$

let $\alpha$ is a angle in triangle where

$\Rightarrow \sin \alpha =\frac{1}{\sqrt{2^2+7^2}}$

$=\frac{1}{\sqrt{50}}$

$=\frac{1}{5\sqrt{2}}$

$\Rightarrow \cos \alpha =\frac{7}{5\sqrt{2}}$

$\sin \beta =\frac{1}{\sqrt{10}},\cos \beta =\frac{\sqrt{10-1}}{\sqrt{10}}=\frac{\sqrt{9}}{\sqrt{10}}=\frac{3}{\sqrt{10}}$

$\sin 2\beta =2\sin \beta \cos \beta =\frac{2\cdot 3}{\sqrt{10}}=\frac{6}{10}$

$\cos 2\beta =\frac{\sqrt{{10}^2-6^2}}{10}=\frac{\sqrt{64}}{10}=\frac{8}{10}$

$\sin (\alpha +2\beta )=\sin \alpha \cos 2\beta +\cos \alpha \sin 2\beta$

$=\frac{1}{5\sqrt{2}}\cdot \frac{8}{10}+\frac{7}{5\sqrt{2}}\cdot \frac{6}{10}$

$=\frac{8+42}{50\cdot \sqrt{2}}=\frac{50}{50\cdot \sqrt{2}}$

$=\frac{1}{\sqrt{2}}$

$\sin (\alpha +2\beta )=\sin \left({45}^o\right)$

$\therefore \alpha +2\beta ={45}^o$