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Question

ABCD is a cyclic quadilateral whose diagonals intersect at a point E. If DBC=70 and BAC=30, find BCD. Further if AB = BC, find ECD.


A

60o

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B

70o

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C

50o

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D

80o

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Solution

The correct option is C

50o


CDB=BAC=30...(i) [Angles in the same segment of a circle are equal] DBC=70.....(ii)
In ΔBCD,
BCD+DBC+CDB=180 [Sum of all the angles of a triangle is 180]
BCD+70+30=180 [Using (i) and (ii)]
BCD+100=180
BCD=180100
BCD=80.....(iii)

In ΔABC, AB=BC
BCA=BAC=30 [Angle opposite to equal sides of a triangle are equal]...(iv)BAC=30 (given)
Now BCD=80
BCA+ECD=80
30+ECD=80
ECD=8030
ECD=50


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