please solve ques 6
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Answer:Q-6angleADC = (b)120°Construction: Join A and D
angleACB=90°[Angle from the diameter forms 90°]
so In ∆ACB,
CAB+ACB+ABC=180°
30+90+ABC=180°
ABC=180°-120°
ABC=60°
In quadrilateral ADCB,
ADC+ABC=180°[Cyclic quadrilateral]
angleADC=180°-120°
angleADC=60°{ANS}
hope it helps u understand
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