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Question
In the given figure, C and D are points on the semi-circle described on AB as diameter. Given angle BAD = 70° and angle DBC = 30°, calculate angle BDC.
Solution
Since ABCD is a cyclic quadrilateral, therefore, ∠BCD + ∠BAD = 180° ...(Since opposite angles of a cyclic quadrilateral are supplementary)
`=>` ∠BCD + 70° = 180°
`=>` ∠BCD = 180° − 70° = 110°
In ΔBCD, we have,
∠CBD + ∠BCD + ∠BDC = 180°
`=>` 30° + 110° + ∠BDC = 180°
`=>` ∠BDC = 180° − 140°
`=>` ∠BDC = 40°
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