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Question
In the given figure, ABCD is a cyclic quadrilateral where diagonals intersect at P such that ∠DBC = 40° and ∠BAC = 60° find ∠BCD
Solution
∠BAC – ∠BDC = 60° ...[Angles at the circumference to the same segment]
∠BCD + ∠BDC + ∠CBD = 180° ...(Sum of the three angles of A is 180°)
∠BCD + 60° + 40° = 180°
∠BCD = 180° – 100°
= 80°
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