Question

The bisectors of any two adjacent angles of a parallelogram intersect at

• A. ${75}^{\circ }$
• B. ${45}^{\circ }$
• C. ${90}^{\circ }$
• D. ${30}^{\circ }$

Answer 1

The correct option is C ${90}^{\circ }$

Let ABCD be parallelogram

$\angle A+\angle B={180}^{\circ }$
[Adjacent angles of parallelogram are supplementary]
Hence, $\frac{\angle A}{2}+\frac{\angle B}{2}=\frac{{180}^{\circ }}{2}={90}^{\circ }$ [Bisectors divide the angles into half]
Thus, $\angle AOB={180}^{\circ }-(\frac{\angle A}{2}+\frac{\angle B}{2})$
$={90}^{\circ }$