Question

In figure, $\angle BAC={60}^{\circ }$ and $\angle BCA={20}^{\circ }$ , find $\angle ADC$.

• A. 15${}^{\circ }$
• B. 50${}^{\circ }$
• C. 80${}^{\circ }$
• D. 40${}^{\circ }$

Answer 1

Answer: (C)

80${}^{\circ }$

In $\Delta ABC$, we have
$\angle BAC+\angle BCA+\angle ABC={180}^{\circ }$....[Angle sum property]
$\Rightarrow {60}^{\circ }+{20}^{\circ }+\angle ABC={180}^o$
$\therefore \angle ABC={180}^{\circ }-{80}^{\circ }={100}^{\circ }$.

$ABCD$ is a cyclic quadrilateral,
$\therefore \angle ABC+\angle ADC={180}^{\circ }$.....[Since, opposite angles of a cyclic quadrilateral are supplementary]
$⟹$ ${100}^{\circ }+\angle ADC={180}^{\circ }$
$⟹\angle ADC={180}^{\circ }-{100}^{\circ }={80}^{\circ }$.

Hence, option $C$ is correct.