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Ex 10.5, 4 - In figure, ∠ABC = 69°, ∠ACB = 31°, find ∠BDC - Angle subtended by arc at the centre

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Ex 10.5, 4 In the given figure, ∠ABC = 69°, ∠ACB = 31°, find ∠BDC. In ΔABC, ∠BAC + ∠ABC + ∠ACB = 180° ∠BAC + 69° + 31° = 180° ∠BAC + 100° = 180° ∠BAC = 180° − 100° ∠BAC = 80° For segment BADCB, ∠ BDC & ∠ BAC are angles in the same segment So, they must be equal ∴ ∠ BDC = ∠ BAC ⇒ ∠BDC = 80°

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Davneet Singh
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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