in figure O is the center of the circle
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D + B = 180° ( opposite angles of cyclic quadrilateral ADCB)
B = 50°
Now ACB = 90° (angle in semicircle)
ACB + B + CAB = 180° (ASP)
CAB = 40°
B = 50°
Now ACB = 90° (angle in semicircle)
ACB + B + CAB = 180° (ASP)
CAB = 40°
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