Question

In the given figure, $\text{PQR}$is a straight line. Find $\text{x}$, then complete the following:​
$\angle \text{AQR}$

Answer 1

Given $\text{PQR}$ is a staright line.
$\Rightarrow \angle \text{PQA = x}+{20}^{\circ }$
$\Rightarrow \angle \text{AQB = 2x}+{10}^{\circ }$
$\Rightarrow \angle \text{BQR =x}-{100}^{\circ }$
Now,
$\because \text{PQR}$ is a straight line we have ,
$\Rightarrow \angle \text{PQR}={180}^{\circ }$
$\Rightarrow \angle \text{(PQA+AQB+BQR)}={180}^{\circ }$[from figure]
$\Rightarrow \text{x}+{20}^{\circ }+2\text{x}+{10}^{\circ }+\text{x}-{10}^{\circ }={180}^{\circ }$
$\Rightarrow 4\text{x}+{20}^{\circ }={180}^{\circ }$
$\Rightarrow 4\text{x}={180}^{\circ }-{20}^{\circ }$
$\Rightarrow \text{x}=\frac{{160}^{\circ }}{4}$
$\therefore \text{x}={40}^{\circ }$
Hence , the value of $\text{x}$ is ${40}^{\circ }$
$\Rightarrow \angle \text{AQR}=\angle \text{AQB}+\angle \text{BQR}$
$\Rightarrow \angle \text{AQR}=3\text{x}$
[since,$\text{x}={40}^{\circ }$]
$\Rightarrow \angle \text{AQR}=3\left({40}^{\circ }\right)$
$\Rightarrow \angle \text{AQR}={120}^{\circ }$
Hence $\angle \text{AQR}={120}^{\circ }$