Question

In the figure, given below, AD = BC,

$\angle BAC={30}^{\circ }$ and

$\angle CBD={70}^{\circ }$. Find $\angle BCD$

• A. ${80}^{\circ }$
• B. ${30}^{\circ }$
• C. ${50}^{\circ }$
• D. ${10}^{\circ }$

Answer 1

The correct option is A

${80}^{\circ }$

In the figure, ABCD is a cyclic quadrilateral AC and BD are its diagonals $\angle BAC={30}^{\circ }$ and $\angle CBD={70}^{\circ }$

$\angle CAD=\angle CBD={70}^{\circ }$

Similarly $\angle BAC=\angle BDC={30}^{\circ }$

$\therefore \angle BAD=\angle BAC+\angle CAD={30}^{\circ }+{70}^{\circ }={100}^{\circ }$

Now $\angle BCD+\angle BAD={180}^{\circ }$

(opposite angles of cyclic quad.)

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

$\Rightarrow \angle BCD={180}^{\circ }-{100}^{\circ }={80}^{\circ }$