in the given figure ABCD is a trapezium in which ab parallel DC ab is a diameter and angle C A B is equals 28 degree the difference of angle ADC and angle abc
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Angle acb =90
(Angle in semicircle)
Therefore angle cba=62
Angle cba +angle cda =180
(Sum of opposite angles of a cyclic quad)
Angle cda=180-62
=118
(Angle in semicircle)
Therefore angle cba=62
Angle cba +angle cda =180
(Sum of opposite angles of a cyclic quad)
Angle cda=180-62
=118
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Answer:
The answer is 56.......
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