Question

In the given figure, PQ is a diameter of a circle with centre O. If $\angle PQR={65}^{\circ },\angle SPR={40}^{\circ }and\angle PQM={50}^{\circ },\text{find}\angle QPR,\angle QPMand\angle PRS$

Answer 1

ANSWER:
Here, PQ is the diameter and the angle in a semicircle is a right angle.
i.e., ∠ PRQ = 90 °In Δ PRQ, we have:
∠ QPR + ∠ PRQ + ∠ PQR = 180° (Angle sum property of a triangle)
⇒ ∠ QPR + 90° + 65° = 180°
⇒ ∠ QPR = (180° – 155°) = 25°
In Δ PQM, PQ is the diameter.
∴ ∠ PMQ = 90°
In Δ PQM, we have:
∠ QPM + ∠ PMQ + ∠ PQM = 180° (Angle sum property of a triangle)
⇒ ∠ QPM + 90° + 50° = 180°
⇒ ∠ QPM = (180° – 140°) = 40°
Now, in quadrilateral PQRS, we have:
∠ QPS + ∠ SRQ = 180° (Opposite angles of a cyclic quadrilateral)
⇒ ∠ QPR + ∠ RPS + ∠ PRQ + ∠ PRS = 180°
⇒ 25° + 40° + 90° + ∠ PRS = 180°
⇒ ∠ PRS = 180° – 155° = 25°
∴ ∠ PRS = 25°
Thus, ∠ QPR = 25°; ∠ QPM = 40°; ∠ PRS = 25°