Question

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Answer 1

Answer: 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°

 

Step-by-step explanation: hope it helps

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