Given, $ABCD$ is a cyclic quadrilateral and $\angle\text{ADC}=140^\circ.$
We know that, sum of the opposite angles in a cyclic quadrilateral is $180^\circ .$
$\angle\text{ADC}+\angle\text{ABC}=180^\circ$
$\Rightarrow\ 140^\circ+\angle\text{ABC}=180^\circ$
$\Rightarrow\ \angle\text{ABC}=180^\circ-140^\circ$
$\therefore\angle\text{ABC}=40^\circ$
Since, $\angle\text{ACB}$ is an angle in a semi-circle.
$\therefore\angle\text{ACB}=90^\circ$
In $\triangle\text{ABC},\ \ \angle\text{BAC}+\angle\text{ACB}+\angle\text{ABC}=180^\circ$ [by angle sum property of a triangle]
$\Rightarrow\angle\text{BAC}+90^\circ+40^\circ=180^\circ$
$\Rightarrow\angle\text{BAC}=180^\circ-130^\circ=50^\circ$
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