Q6 of 141 Page 438

In the given figure, and find

BD = CD


∴∠CBD = ∠BCD = 30°


In triangle BCD,


∠BDC + ∠BCD + ∠CBD = 180°[Sum of angles of triangle]


⇒ ∠BDC + 30° + 30° = 180°


⇒ ∠BDC + 60° = 180°


⇒ ∠BDC = 120°


Now,


∠BDC + ∠BAC = 180°[ABCD is a cyclic quadrilateral]


⇒ 120° + ∠BAC = 180°


⇒ ∠BAC = 60°


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