Question

Question 131

In $\Delta PQR$, if $3\angle P=4\angle Q=6\angle R$, calculate the angles of the triangle.

Answer 1

Given, $3\angle P=4\angle Q=6\angle R$
Then, $\angle P=\frac{6}{3}\angle R=2\angle R\angle Q=\frac{6}{4}\angle R=\frac{3}{2}\angle R$

In $\Delta PQR$,
$\angle P+\angle Q+\angle R={180}^{\circ }$ [angle sum property of a triangle]
$\Rightarrow 2\angle R+\frac{3}{2}\angle R+\angle R={180}^{\circ }\Rightarrow 3\angle R+\frac{3}{2}\angle R={180}^{\circ }$
$\Rightarrow 6\angle R+3\angle R={180}^{\circ }\times 2$ [on taking LCM in LHS]
$\Rightarrow 9\angle R={360}^{\circ }\Rightarrow \angle R=\frac{{360}^{\circ }}{9}={40}^{\circ }\therefore \angle P=2\angle R=2\times {40}^{\circ }={80}^{\circ }$
and $\angle Q=\frac{3}{2}\angle R=\frac{3}{2}\times {40}^{\circ }={60}^{\circ }$
Hence, all the angles of the triangle are ${80}^{\circ },{60}^{\circ }$ and ${40}^{\circ }$