Question

please solve ques 6​

Answer 1

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