Question

In the given figure, PQR is a straight line. Find 𝑥, then complete the following:​
(i) $\angle AQB=$

Answer 1

Given, $PQR$ is a straight line

$\angle PQA=x+{20}^{\circ }$

$\angle AQB=2x+{10}^{\circ }$

$\angle BQR=x-{10}^{\circ }$

Now,
$\therefore$ $PQR$ is a straight line we have,

$\angle PQR={180}^{\circ }$

$\angle PQR+\angle AQB+\angle BQR={180}^{\circ }$
[from figure]


$x+{20}^{\circ }+2x+{10}^{\circ }+x-{10}^{\circ }={180}^{\circ }$


$4x+{20}^{\circ }={180}^{\circ }$

$4x={180}^{\circ }-{20}^{\circ }$

$x=\frac{{160}^{\circ }}{4}={40}^{\circ }$

Hence, the value of $x$ is ${40}^{\circ }$.

$\angle AQB=2x+{10}^{\circ }$

$\angle AQB=2\times {40}^{\circ }+{10}^{\circ }$

$\angle AQB={80}^{\circ }+{10}^{\circ }$

$\therefore$ $\angle AQB={90}^{\circ }$

Hence , $\angle AQB={90}^{\circ }$