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Ex 9.3, 5 In the given figure, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠BEC = 130° and ∠ECD = 20°. Find ∠BAC. Since BD is a line ∠ BEC + ∠ DEC = 180° 130° + ∠ DEC = 180° ∠ DEC = 180° – 130° ∠ DEC = 50° In Δ DEC ∠ DEC + ∠ EDC + ∠ ECD = 180° 50° + ∠ EDC + 20° = 180° 70° + ∠ EDC = 180° ∠ EDC = 180° – 70° ∠ EDC = 110° For segment BADCB, ∠ BAC & ∠ BDC are in the same segment So, they must be equal ∴ ∠ BAC = ∠ BDC ∠ BAC = 110°

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo