Question

If $\alpha and\beta$ are the angles in the first Quadrant, $\tan \alpha =\frac{1}{7},\sin \beta =\frac{1}{\sqrt{10}}$ then the value of $\alpha +2\beta$ is

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

Answer 1

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

$\Rightarrow \tan \alpha =\frac{1}{7}$

$\Rightarrow \sin \beta =\frac{1}{\sqrt{14}}$

In right angled $△ABC,$

$\Rightarrow A{B}^2+B{C}^2=A{C}^2$

$\Rightarrow B{C}^2=A{C}^2-A{B}^2$

$={\left(\sqrt{10}\right)}^2-{\left(\sqrt{1}\right)}^2$

$=10-1$

$\Rightarrow BC=\sqrt{9}=3$

Now, $\tan \beta =\frac{AB}{BC}=\frac{1}{3}$

$\&\tan 2\beta =\frac{2\tan \beta }{1-{\tan }^2\beta }=\frac{2\times \frac{1}{3}}{1-\frac{1}{9}}=\frac{\frac{2}{3}}{\frac{8}{9}}$

$=\frac{2}{3}\times \frac{9}{8}$

$=\frac{3}{4}$

Now, $\tan (\alpha +2\beta )=\frac{\tan \alpha +\tan 2\beta }{1-\tan \alpha \tan 2\beta }$

$=\frac{\frac{1}{7}+\frac{3}{4}}{1-\frac{1}{7}\frac{3}{4}}$

$=\frac{\frac{25}{28}}{\frac{28-3}{28}}$

$=\frac{25}{28}\times \frac{28}{25}$

$=1$

$\Rightarrow \tan (\alpha +2\beta )=\tan 45°$

$\Rightarrow \alpha +2\beta =45°$

Hence, the answer is $45°.$