Sum
In the given figure, ∠POQ = 100° and ∠PQR° = 30°, then find ∠RPO
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Solution
∠PRQ = `1/2 ∠"POQ"` ...(Angles at the circumference)
= `1/2 xx 100^circ`
= 50°
∠OPQ + ∠OQP = 180° – 100° ...(Total angles of A is 180°)
= 80°
∴ ∠OPQ = `80/2`
= 40° ...(Since OP = OQ, radius of the circle)
∠RPQ = 180° − (30 + 50)°
= 100°
∴ ∠RPO = ∠RPQ – ∠OPQ
= 100° – 40°
= 60°
Concept: Theorem: If One Side of a Cyclic Quadrilateral is Produced Then the Exterior Angle is Equal to the Interior Opposite Angle.
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