Mathematics
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.
Answer
As ABCD is a cyclic quadrilateral, we have
⇒ ∠BCD + ∠BAD = 180° [Opposite angles of a cyclic quadrilateral are supplementary]
⇒ ∠BCD + 70° = 180°
⇒ ∠BCD = 180° - 70° = 110°
In ∆BCD,
⇒ ∠CBD + ∠BCD + ∠BDC = 180° [By angle sum property of triangle]
⇒ 30° + 110° + ∠BDC = 180°
⇒ ∠BDC = 180° - 140° = 40°.
Hence, ∠BDC = 40°.
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