Question

In the given figure, $AOB$ is a straight line . Find the value of $x$ and also answer the following.(iv) Which angle is actue?

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



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

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

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

Hence, the value of $x$ is ${90}^{\circ }$.

Now,
$\angle AOP={120}^{\circ }$ and $\angle BOP={60}^{\circ }$

And we know that acute angle is less than ${90}^{\circ }$

$\therefore$ $\angle BOP$ is an acute angle

Hence, $\angle BOP$ is an acute angle.