Question

In a $△PQR$ given that $x:y:z::2:3:6,$ then the value of $\angle QPR$ (write the ans in degrees )

Answer 1

$\tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan beta}$
$=\frac{\frac{y}{z}+\frac{x}{z}}{1-\frac{y}{z}.\frac{x}{z}}$
$=\frac{\frac{3}{6}+\frac{2}{6}}{1-\frac{3}{6}.\frac{2}{6}}$
$=1$ on simplification
$\Rightarrow \tan (\alpha +\beta )=1$
$\Rightarrow \angle QPR=\alpha +\beta =\frac{\pi }{4}$