Question

$\text{(A) In the given figure,}\left(p\right){60}^{\circ }\text{ABCD is a cyclic Quadrilateral.}\text{O is the centre of the circle.}\text{If}\angle BOD={160}^{\circ }\text{find the Measure of}\angle BPD$

• A. ${80}^{\circ }$
• B. ${100}^{\circ }$
• C. ${120}^{\circ }$
• D. ${130}^{\circ }$

Answer 1

The correct option is B

${100}^{\circ }$

Consider the arc BCD of the circle. This arc makes angle $\angle BOD={160}^{\circ }$ at the centre of the circle and $\angle BAD$ at a point A on the remaining circumference.



$\therefore \angle BAD=\frac{1}{2}\angle BOD={80}^{\circ }$
Now, ABPD is a cyclic quadrilateral.

$\because \text{Sum of opposite angles in a cyclic quadrilateral}={180}^{\circ }$
$\Rightarrow \angle BAD+\angle BPD={180}^{\circ }$
$\Rightarrow {80}^{\circ }+\angle BPD={180}^{\circ }\Rightarrow \angle BPD={100}^{\circ }$