Question

$ABCD$ is a cyclic quadrilateral in a circle with centre $O$. If $\angle ADC={130}^{\circ }$, find $\angle BAC$.

Answer 1

We observe that

$\angle ACB={90}^{\circ }$ [Angle in a semi-circle is ${90}^{\circ }$]
Also,
$\angle ABC={180}^{\circ }–\angle ADC$

[Pair of opposite angles in a cyclic quadrilateral are supplementary]

$={180}^{\circ }–{130}^{\circ }$

$={50}^{\circ }$
By angle sum property of the right-triangle $ABC,$

we have $\angle ABC+\angle ACB+\angle BAC={180}^{\circ }$

$\Rightarrow \angle BAC={180}^{\circ }-{90}^{\circ }-\angle ABC$

$\Rightarrow \angle BAC={90}^{\circ }–\angle ABC$
$={90}^{\circ }–{50}^{\circ }$

Thus, $\angle BAC={40}^{\circ }$