Question

In the given figure, ${60}^{\circ }$
ABCD is a cyclic quadrilateral.
O is the center of the circle.
If $\angle BOD={160}^{\circ }findtheMeasureof\angle BPD.$

• A. ${60}^{\circ }$
• B. ${100}^{\circ }$
• C. ${90}^{\circ }$
• D. ${120}^{\circ }$

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 center of the circle and $\angle BAD$ at a point A on the circumference.

$\therefore \angle BAD=\frac{1}{2}\angle BOD=\frac{{160}^{\circ }}{2}={80}^{\circ }$ ($\because$Angle subtended by an arc of a circle at the centre of the circle is double the angle subtended by the arc on the circumference.)
Now, ABPD is a cyclic quadrilateral.

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