Question

ABCD is a cyclic quadilateral whose diagonals intersect at a point E. If $\angle DBC={70}^{\circ }$ and $\angle BAC={30}^{\circ }$, find $\angle BCD$. Further if $AB=BC,$ find $\angle ECD$.

• A. ${60}^{\circ },{20}^{\circ }$
• B. ${70}^{\circ },{40}^{\circ }$
• C. ${80}^{\circ },{50}^{\circ }$
• D. ${80}^{\circ },{30}^{\circ }$

Answer 1

The correct option is C

${80}^{\circ },{50}^{\circ }$

$\angle CDB=\angle BAC={30}^{\circ }$ ...(i) [Angles in the same segment of a circle are equal]

$\angle DBC={70}^{\circ }$ .....(ii)
In $△BCD,$
$\angle BCD+\angle DBC+\angle CDB={180}^{\circ }$ [Sum of all the angles of a triangle is ${180}^{\circ }$]
$\angle BCD+{70}^{\circ }+{30}^{\circ }={180}^{\circ }$ [Using (i) and (ii)]
$\angle BCD+{100}^{\circ }={180}^{\circ }$
$\angle BCD={180}^{\circ }-{100}^{\circ }$
$\angle BCD={80}^{\circ }$ .....(iii)

In $△ABC,AB=BC$
$\therefore \angle BCA=\angle BAC={30}^{\circ }$ .....(iv) [Angle opposite to equal sides of a triangle are equal]

$\angle BAC={30}^{\circ }$ [given]
Now, $\angle BCD={80}^{\circ }$
$\Rightarrow \angle BCA+\angle ECD={80}^{\circ }$
$\Rightarrow {30}^{\circ }+\angle ECD={80}^{\circ }$
$\Rightarrow \angle ECD={80}^{\circ }-{30}^{\circ }$
$\Rightarrow \angle ECD={50}^{\circ }$