Question

In the figure given figure, $AOB$ is a straight line. Find the value of $x$ and also answer the following:
ii) $\angle BOP=$

Answer 1

$AOB$ is straight line.
$\therefore \angle AOB={180}^{\circ }$
$\angle AOP+\angle BOP={180}^{\circ }$
$x+{30}^{\circ }+x-{30}^{\circ }={180}^{\circ }$
$[\text{Given},\angle AOP=x+{30}^{\circ }\text{and}\angle BOP=x-{30}^{\circ }]$
$2x={180}^{\circ }$
$x=\frac{{180}^{\circ }}{2}$
$\therefore x={90}^{\circ }$
Hence, the value of $x$ is ${90}^{\circ }$.

$\angle BOP=x-{30}^{\circ }$
$\angle BOP={90}^{\circ }-{30}^{\circ }[\because x={90}^{\circ }]$
$\Rightarrow \angle BOP={60}^{\circ }$
Hence, $\angle BOP\text{is}{60}^{\circ }$.