Question

In the given circle, AD=BC, $\angle BAC={30}^o$ and $\angle CBD={70}^o$. Find $\angle BCD$.

• A. ${80}^0$
• B. ${30}^0$
• C. ${70}^0$
• D. None of these

Answer 1

The correct option is A ${80}^0$

In figure, $ABCD$ is cyclic quadrilateral.

Here, $AC$ and $BD$ are diagonals.

Also, $\angle BAC={30}^{\circ }$ and $\angle CBD={70}^{\circ }$.

Now, $\angle BAC=\angle BDC={30}^{\circ }$.

Then, $\angle BAD=\angle BDC+\angle CAD$

$={30}^{\circ }+{70}^{\circ }={100}^{\circ }$.

Now, $\angle BCD+\angle BAD={180}^{\circ }$ ...(opposite angles are cyclic quadrilateral)

$\Rightarrow \angle BCD+{100}^{\circ }={180}^{\circ }$

$\Rightarrow \angle BCD={180}^{\circ }-{100}^{\circ }$

$\Rightarrow \angle BCD={80}^{\circ }$.

Therefore, option $A$ is correct.