Question

ABCD is a cyclic quadrilateral angle D=61 and a diagonal pass through A and C and angle BAC =angle BCA find angle BAC
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Answer 1

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Given :- angle BAC = 50° and angle DBC = 60°

To find :- angle BCD.

Solution :-

angle BAC = angle CDB [Angles on the same segment of a circle are equal]

angle CDB = 50°

In triangle BCD,

angle BCD + angle CDB + angle DBC = 180° [Angle sum property of a triangle]

angle BCD + 50° + 60° = 180°

angle BCD + 110° = 180°

angle BCD = 180° - 110°

angle BCD = 70° ANSWER

Answer 2

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Given :- angle BAC = 50° and angle DBC = 60°

To find :- angle BCD.

Solution :-

angle BAC = angle CDB [Angles on the same segment of a circle are equal]

angle CDB = 50°

In triangle BCD,

angle BCD + angle CDB + angle DBC = 180° [Angle sum property of a triangle]

angle BCD + 50° + 60° = 180°

angle BCD + 110° = 180°

angle BCD = 180° - 110°

angle BCD = 70° ANSWER

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