In a cyclic quadrilateral PQRS,angle P = angl | vedclass

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Question

(A) In a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$.Therefore,$\angle P + \angle R = 180^{\circ}$.Given that $\angle P = \angle

Vedclass · Accepted Answer

$\angle P = 115^{\circ}, \angle R = 65^{\circ}$

Vedclass · Answer

(A) In a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$.Therefore,$\angle P + \angle R = 180^{\circ}$.Given that $\angle P = \angle R + 50^{\circ}$.Substituting the value of $\angle P$ in the first equation:$(\angle R + 50^{\circ}) + \angle R = 180^{\circ}$$2\angle R + 50^{\circ} = 180^{\circ}$$2\angle R = 130^{\circ}$$\angle R = 65^{\circ}$.Now,finding $\angle P$:$\angle P = 65^{\circ} + 50^{\circ} = 115^{\circ}$.Thus,$\angle P = 115^{\circ}$ and $\angle R = 65^{\circ}$.

In a cyclic quadrilateral $PQRS$,$\angle P = \angle R + 50^{\circ}$. Find $\angle P$ and $\angle R$.

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