Question

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

Answer 1

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

Now,
$\angle AOP=x+{30}^{\circ }$
$\angle AOP={90}^{\circ }+{30}^{\circ }[\because x={90}^{\circ }]$
$\therefore \angle AOP={120}^{\circ }$
Hence, $\angle AOP\text{is}{120}^{\circ }$.