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Question

ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and if ADC=140o, then BAC is equal to:

A
60o
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B
50o
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C
40o
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D
30o
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Solution

The correct option is B 50o
Given:
ABCD is a cyclic quadrilateral with AB as the diameter of the circle.
Also, ADC=140o.
We have to find: BAC.

Then,
ADC+ABC=180o ....(since the sum of the opposite angles of a quadrilateral is 180o).
ABC=180o140o=40o.
Also, ACB=90o ....(since the angle subtended by a diameter at the circumference of the circle, is 90o).
In ΔABC we have,
ACB=90o and ABC=40o.
So, CAB=180o(ACB+ABC)
=180o(90o+40o)
BAC=50o .

Hence, option B is correct.

207058_77949_ans_5a7ff7539a0b45d8b21bb9936d9f1ed3.jpg

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