#### 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°

Is there an error in this question or solution?

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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. Concept: Cyclic Properties.

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