Question

10. In the following figure AB is a diameter of the circle. If <ADC = 120O

Find angle < CBA and < BAC​

Answer 1

CBA+ADC=180(Since ABCD is a cyclic quad.)

CBA= 180-ADC

=180-120

=60 ans.

NOW,in ∆ACB,

ACB=90(angle subtended by semicircle)

according to Angle Sum prop.of a ∆

ACB+ABC+CAB=180

90+60+CAB=180

150+CAB=180

CAB=180-150

CAB=30ans.