In the given figure, PQ is a diameter of a circle with centre O.
Question:

In the given figure, PQ is a diameter of a circle with centre O. If PQR = 65°, ∠SPR = 40° and ∠PQM = 50°, find ∠QPR, ∠QPM and ∠PRS.

 

Solution:

Here, PQ is the diameter and the angle in a semicircle is a right angle.
i.e., PRQ = 90°
In ΔPRQ, we have:
QPR + PRQ + PQR = 180°   (Angle sum property of a triangle)
⇒ QPR + 90° + 65° = 180°
 ⇒QPR = (180° – 155°) = 25°

In ΔPQM, PQ is the diameter.
PMQ = 90°
In ΔPQM, 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°

 

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