Question

In the figure $\text{PQR}$ is a straight line .Find $x$ and complete the following .
$\angle \text{PBQ}=$

Answer 1

Given $\text{PQR}$ is a straight line
$\Rightarrow \angle \text{PQA}=\text{x}+{20}^{\circ }$
$\Rightarrow \angle \text{AQB}=2\text{x}+{10}^{\circ }$
$\Rightarrow \angle \text{BQR}=\text{x}-{10}^{\circ }$

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 x is ${40}^{\circ }$
$\Rightarrow \angle \text{BQP}=\angle \text{PQA}+\angle \text{AQB}$
$\Rightarrow \angle \text{BQP= 3x}+{30}^{\circ }$
$\Rightarrow \angle \text{BQP}=3\times {40}^{\circ }+{30}^{\circ }[Since,x={90}^{\circ }$]
$\Rightarrow \angle \text{BQP}={150}^{\circ }$
Hence $\angle \text{BQP}={150}^{\circ }$