Question
In Fig. if $AOB$ is a diameter and $\angle\text{ADC}=120^\circ,$ then $\angle\text{CAB}=30^\circ.$ 
✓
Answer
Join $CA$ and $CB.$
Since, $ADCB$ is a cyclic quadrilateral.
$\angle\text{ADC}+\angle\text{CBA}=180^\circ [$sum of opposite angles of cyclic quadrilateral is $180^\circ ]$
$\Rightarrow\angle\text{CBA}=180^\circ-120^\circ=60^\circ\ \ [\therefore\angle\text{ADC}=120^\circ]$
In $\triangle\text{ACB,}\ \angle\text{CAB} + \angle\text{CBA} + \angle\text{ACB} = 180^\circ$
[by angle sum property of a triangle]
$\angle\text{CAB}+60^\circ+90^\circ=180^\circ$
$\big[$triangle formed from diameter to the circle is $90^\circ$
i.e., $\angle\text{ACB}=90^\circ\big]$
$\Rightarrow\angle\text{CAB}=180^\circ-150^\circ=30^\circ.$
Since, $ADCB$ is a cyclic quadrilateral.
$\angle\text{ADC}+\angle\text{CBA}=180^\circ [$sum of opposite angles of cyclic quadrilateral is $180^\circ ]$
$\Rightarrow\angle\text{CBA}=180^\circ-120^\circ=60^\circ\ \ [\therefore\angle\text{ADC}=120^\circ]$
In $\triangle\text{ACB,}\ \angle\text{CAB} + \angle\text{CBA} + \angle\text{ACB} = 180^\circ$
[by angle sum property of a triangle]
$\angle\text{CAB}+60^\circ+90^\circ=180^\circ$
$\big[$triangle formed from diameter to the circle is $90^\circ$
i.e., $\angle\text{ACB}=90^\circ\big]$
$\Rightarrow\angle\text{CAB}=180^\circ-150^\circ=30^\circ.$
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