Question

In the given figure, POQ is a diameter and PQRS is a cyclic quadrilateral. If $\angle PSR={150}^{\circ },\text{find}\angle RPQ$.

Answer 1

ANSWER:
In cyclic quadrilateral PQRS, we have:
∠ PSR + ∠ PQR = 180 °
⇒ 150 ° + ∠ PQR = 180 °
⇒ ∠ PQR = (180 ° – 150 ° ) = 30 °
∴ ∠ PQR = 30 °
...(i)
Also, ∠ PRQ = 90 ° (Angle in a semicircle)
Now, in Δ PRQ, we have:
...(ii)∠ PQR + ∠ PRQ + ∠ RPQ = 180 °
⇒ 30 ° + 90 ° + ∠ RPQ = 180 ° [From(i) and (ii)]
⇒ ∠ RPQ = 180 ° – 120 ° = 60 °
∴ ∠ RPQ = 60 °