Question

In $△PQR\angle P=x^{\circ },\angle Q=3x^{\circ },\angle R=y^{\circ }$. If $3y-5x=30$, then find the value of each angle of $△PQR$.

Answer 1

Given,
In $△PQR,\angle P=x^{\circ },\angle Q=3x^{\circ },\angle R=y^{\circ }$
We know that, $\angle P+\angle Q+\angle r={180}^{\circ }$
$\Rightarrow x+3x+y={180}^{\circ }$
$\Rightarrow 4x+y={180}^{\circ }....\left(1\right)$
By question $3y-5x={30}^{\circ }.....\left(2\right)$
Multiply equation $\left(1\right)$ by $3$ and then subtracting equation $\left(2\right)$,
$(4x+y={180}^{\circ })\times 3$
$\Rightarrow 12x+3y={540}^{\circ }$
$\underset{–}{\underset{+}{-}5x+\underset{-}{3}y=\underset{-}{{30}^{\circ }}}$
$\Rightarrow 17x={510}^{\circ }$
$\Rightarrow x=\frac{510}{17}={30}^{\circ }$
Putting value of $x$ in equation $\left(1\right)$,
$\Rightarrow 4x+y={180}^{\circ }$
$\Rightarrow 4\times 30+y={180}^{\circ }$
$\Rightarrow 120+y={180}^{\circ }$
$\Rightarrow y={180}^{\circ }-{120}^{\circ }$
$\Rightarrow y={60}^{\circ }$
$\therefore \angle P=x^{\circ }={30}^{\circ }$
$\angle Q=3x^{\circ }=3\times {30}^{\circ }={90}^{\circ }$
and $\angle R=y^{\circ }={60}^{\circ }$
Hence, angles of $△PQR$ are ${30}^{\circ },{60}^{\circ }$ and ${90}^{\circ }$.