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

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Solution

$We know that the angle in a sem - circle is 90 ° . So, ∠ Q R P = ∠ P M Q = 90 ° (i) In ∆ Q P R, ∠ Q P R + ∠ Q R P + ∠ P Q R = 180 ° (A ngle - sum property) \Rightarrow ∠ Q P R = 180 ° - ∠ Q R P - ∠ P Q R \Rightarrow ∠ Q P R = 180 ° - 90 ° - 65 ° \Rightarrow ∠ Q P R = 25 ° Hence, ∠ Q P R = 25 ° .$

$(i) In ∆ Q P M, ∠ Q P M + ∠ Q M P + ∠ P Q M = 180 ° (A ngle - sum property) \Rightarrow ∠ Q P M = 180 ° - ∠ Q M P - ∠ P Q M \Rightarrow ∠ Q P M = 180 ° - 90 ° - 50 ° \Rightarrow ∠ Q P M = 40 ° Hence, ∠ Q P M = 40 ° .$

$(iii) P Q R S is a cyclic quadrilateral . S o, ∠ S P Q + ∠ Q R S = 180 ° . \Rightarrow ∠ Q P R + ∠ S P R + ∠ Q R P + ∠ P R S = 180 ° \Rightarrow ∠ P R S = 180 ° - ∠ Q P R - ∠ S P R - ∠ Q R P \Rightarrow ∠ P R S = 180 ° - 25 ° - 40 ° - 90 ° \Rightarrow ∠ P R S = 25 ° Hence, ∠ P R S = 25 ° .$

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