Question

In $\Delta PQR$, if $2\angle P=3\angle Q=6\angle R$. calculate measures of each of the angles.

Answer 1

Let $2\angle P=3\angle Q=6\angle R=x$
Then $\angle P=\frac{x}{2},\angle Q=\frac{x}{3},\angle R=\frac{x}{6}$
Now, $\angle P+\angle Q+\angle R={180}^{\circ }$(angle sum property)
$\therefore \frac{x}{2}+\frac{x}{3}+\frac{x}{6}={180}^{\circ }$
$\Rightarrow \frac{3x+2x+x}{6}={180}^{\circ }$
$\Rightarrow \frac{6x}{6}={180}^{\circ }\Rightarrow x={180}^{\circ }$
$\therefore$ Angles are $\angle P=\frac{{180}^{\circ }}{2}={90}^{\circ }$
$\angle Q=\frac{{180}^{\circ }}{3}={60}^{\circ }$ and $\angle R=\frac{{180}^{\circ }}{6}={30}^{\circ }$