Question

$ABCD$ is a quadrilateral inscribed in a circle. Diagonals $AC$ and $BD$ are joined.
If $\angle CAD={60}^{\circ }$ and $\angle BDC={25}^{\circ }$. Find $\angle BCD$

• A. ${85}^{\circ }$
• B. ${120}^{\circ }$
• C. ${115}^{\circ }$
• D. ${95}^{\circ }$

Answer 1

The correct option is D ${95}^{\circ }$

$\angle CAD={60}^{\circ }$

& $\angle BDC={25}^{\circ }$

$\angle CAD=\angle DBC={60}^{\circ }$ [As angles are in same segment].

In $△BCD$,

$\angle DBC+\angle BDC+\angle BCD={180}^{\circ }$

$\Rightarrow {60}^{\circ }+{25}^{\circ }+\angle BCD={180}^{\circ }$

$\Rightarrow \angle BCD={180}^{\circ }-{85}^{\circ }={95}^{\circ }$