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Ex 10.5, 3 - In figure, ∠PQR = 100°, where P, Q and R - Cyclic quadrilaterals

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  1. Chapter 10 Class 9 Circles
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Ex 10.5, 3 In the given figure, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR. Here, PR is chord We mark point S on the major arc of the circle. ∴ PQRS is a cyclic quadrilateral. So, ∠PQR + ∠PSR = 180° 100° + ∠PSR = 180° ∠PSR = 180° − 100° ⇒ ∠PSR = 80° Arc PQR subtends ∠ POR at centre of circle. And ∠ PSR on point S So, ∠ POR = 2 ∠ PSR ∠POR = 2 × (80°) = 160° Now, In ∆ OPR OP = OR ∴ ∠OPR = ∠ORP Also, in ∆ OPR ∠OPR + ∠ORP + ∠POR = 180° ∠OPR + ∠OPR + ∠POR = 180° 2 ∠OPR + 160° = 180° 2 ∠OPR = 180° − 160° 2 ∠OPR = 20° ∠OPR = (20°)/2 ∠OPR = 10°

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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