Circles — Maths STD 9 — Question

We know that the opposite angles of a quadrilateral are supplementary.
$\angle\text{ADC}+\angle\text{ABC}=180^\circ$
$\Rightarrow\ 120^\circ+\angle\text{ABC}=180^\circ$
$\Rightarrow\ \angle\text{ABC}=60^\circ$
Since BOC is a diameter $\angle\text{ACB}=90^\circ.$
In $\triangle\text{CAB},$
$\angle\text{ABC}+\angle\text{BAC}+\angle\text{ACB}=180^\circ$ [Angle sum property]
$\Rightarrow\ 60^\circ+\angle\text{BAC}+90^\circ=180^\circ$
$\Rightarrow\ \angle\text{BAC}=30^\circ$