Question

In figure,$A,B,C$ and $D$ are four points on a circle. $AC$ and $BD$ intersect at a point $E$ such that $\angle BEC={130}^{\circ }$ and $\angle ECD={20}^{\circ }$. Find $\angle BAC$ (in degrees).

Answer 1

$\angle CED+\angle BEC={180}^{\circ }$ (Linear Pair)

$\Rightarrow \angle CED+{130}^{\circ }={180}^{\circ }$

$\Rightarrow \angle CED+{180}^{\circ }-{130}^{\circ }={50}^{\circ }$ ...............(i)

Now, $\angle ECD={20}^0$ ...............(ii)

In $\Delta CED$,

$\angle CED+\angle ECD+\angle CDE={180}^0$ [Sum of all the angles of a triangle is ${180}^{\circ }$]

$\Rightarrow {50}^{\circ }+{20}^{\circ }+\angle CDE={180}^{\circ }$ [Using (i) and (ii)]

$\Rightarrow {70}^{\circ }+\angle CDE={180}^{\circ }$

$\Rightarrow \angle CDE={180}^{\circ }-{70}^{\circ }$

$\Rightarrow \angle CDE=\angle CDB={110}^{\circ }$ .............(iii)

Now $\angle BAC=\angle CDB={110}^{\circ }$ (angles in the same segment are equal)

Hence, $\angle BAC={110}^{\circ }$