Question

CAN SOMEONE PLEASE HELP ME !!!!
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Answer 1

(i) ∠QPR

∴ PQ is a diameter

∴ ∠PRQ = 900 [Angle in a semi-circle is 900]

In ΔPQR,

∠QPR + ∠PRQ + ∠PQR = 1800 [Angle Sum Property of a triangle]

=> ∠QPR + 900 + 650= 1800

=> ∠QPR + 1550 = 1800

=> ∠QPR = 1800 - 1550

=> ∠QPR = 250.

(ii) ∠PRS

∴ PQRS is a cyclic quadrilateral

∴ ∠PSR + ∠PQR = 1800 [∴ Opposite angles of a cyclic quadrilateral are supplementary]

=> ∠PSR + 650 = 1800

=> ∠PSR = 1800- 650

=> ∠PSR = 1150

In DPSR,

∠PSR + ∠SPR + ∠PRS = 1800 [Angles Sum Property of a triangle]

=> 1150 + 400 + ∠PRS = 1800

=> 1150 + ∠PRS = 1800

=> ∠PRS = 1800 - 1550

=> ∠PRS = 250

(iii) ∠QPM

∴ PQ is a diameter

∴ ∠PMQ = 900 [∴ Angle in a semi - circle is 900]

In ΔPMQ,

∠PMQ + ∠PQM + ∠QPM = 1800 [Angle sum Property of a triangle]

=> 900 + 500 + ∠QPM = 1800

=> 1400 + ∠QPM = 1800

=> ∠QPM = 1800 - 1400

=> ∠QPM = 400.

hope it works

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