Question

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

Answer 1

$AOB$ is astraight line.

$\therefore$ $\angle AOB={180}^{\circ }$

$\angle AOP+\angle BOP={180}^{\circ }$

$x+{30}^{\circ }+x-{30}^{\circ }={180}^{\circ }$

[Given ,$\angle AOP=x+{30}^{\circ }$ and BOP = x - 30]


$2x={180}^{\circ }$

$x=\frac{{180}^{\circ }}{2}={90}^{\circ }$

$\therefore$ $x={90}^{\circ }$


$\angle BOP=x-{30}^{\circ }$

$\angle BOP={90}^{\circ }-{30}^{\circ }$ [since, $x={90}^{\circ }$]

$\therefore$ $\angle BOP={60}^{\circ }$

Hence, $\angle BOP$ is ${60}^{\circ }$.