Question

In the given figure, O is the centre of the circle. If Reflex
POR = 210° and ORQ = 60°, find: (i) RPQ (ii) PQO
(11) ORP (iv) PQR
​

Answer 1

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Answer 2

Answer :-

→ ∠POR = 360° - 210° = 150° .

so,

→ Angle at circumference by chord PR = (150/2) = 75° .

then,

→ ∠PQR = 180° - 75° = 105° (Ans.iv)

now, in quadrilateral PQRO,

→ ∠POR + ∠ORQ + ∠RQP + ∠QPO = 360° (By angle sum property.)

→ 150° + 60° + 105° + ∠QPO = 360°

→ 315° + ∠QPO = 360°

→ ∠QPO = 360° - 315° = 45° .

now, in ∆POR, we have,

→ ∠POR = 150°

→ OP = OR (radius)

so,

→ ∠OPR = ∠ORP (Angle opposite to equal sides are equal.)

→ ∠OPR = ∠ORP = (180° - 150°)/2 = 30/2 = 15° (Ans.iii) .

then,

→ ∠RPQ = ∠QPO - ∠OPR

→ ∠RPQ = 45° - 15° = 30° (Ans.i)

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