In the given figure ABCD is a cyclic quadilateral; O is the centre of the circle. If $∠$ BOD= $160^{\circ}$ ,find the measure of $∠$ BPD= $x^{\circ}$ .The value of x is

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Solution

Consider the arc BCD of the circle. This arc makes $∠$ BOD = $160^{\circ}$ at the centre of the circle and $∠$ BAD at a point A on the circumference.
$∠$ BAD = $∠$ BOD = $80^{\circ}$
Now, ABPD is a cyclic quadrilateral.
$∠$ BAD + $∠$ BPD = $180^{\circ}$
$80^{\circ}$ + $∠$ BPD = $180^{\circ}$
$∠$ BPD = $100^{\circ}$
$∠$ BCD = $100^{\circ}$ [ $∵$ $∠$ BPD and $∠$ BCD are angles in the same segment $∴$ $∠$ BCD = $∠$ BPD ]

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∠ B O D = 160^{\circ}

∠ B P D

Consider the cyclic quadrilateral ABCD. If O is the centre of the circle and $∠$ BOD = $160^{\circ}$ , then find the measure of $∠$ BPD.

∠

^{\circ}

∠

In the given figure, ABCD is a cyclic Quadrilateral. O is the centre of the circle and $∠ B O D = 160^{\circ}$ , find the measure of $∠ B P D .$

\begin{matrix} C o l u m n - I & C o l u m n - I I (A) I n t h e g i v e n f i g u r e, & (p) 60^{\circ} A B C D i s a c y c l i c Q u a d r i l a t e r a l . O i s t h e c e n t r e o f t h e c i r c l e . I f ∠ B O D = 160^{\circ} f i n d t h e M e a s u r e o f ∠ B P D . \end{matrix}

\begin{matrix} (B) I n g i v e n f i g u r e, A B C D i s a c y c l i c & (q) 65^{\circ} q u a d r i l a t e r a l w h o s e s i d e A B i s a d i a m e t e r o f t h e c i r c l e t h r o u g h A, B, C, D . I f ∠ A D C = 130^{\circ}, f i n d ∠ B A C . \end{matrix}

\begin{matrix} (C) I n t h e g i v e n f i g u r e B D = D C a n d & (r) 40^{\circ} ∠ C B D = 30^{\circ} F i n d m (∠ B A C) \end{matrix}

\begin{matrix} (D) I n t h e g i v e n f i g u r e, O i s t h e c e n t r e o f t h e & (s) 100^{\circ} a r c A B C s u b t e n d s a n a n g l e o f 130^{\circ} a t t h e c e n t r e . I f A B i s e x t e n d e d t o P m f i n d ∠ P B C . \end{matrix}

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