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

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Given:

A, B, C and D are four points on a circle. $\mathrm{AC}$ and $\mathrm{BD}$ intersect at a point $\mathrm{E}$ such that $\angle \mathrm{BEC}=130^{\circ}$ and $\angle \mathrm{ECD}=20^{\circ}$.
To do:

We have to find $\angle \mathrm{BAC}$.

Solution:

We know that,

The angles in the segment of a circle are equal.

This implies,

$\angle BAC = \angle CDE$

$\angle CEB = \angle CDE+\angle ECD$      (Exterior angle property)

$130^o=\angle CDE+20^o$

$\angle CDE=130^o-20^o$

$=110^o$

Therefore,

$\angle BAC= CDE =110^o$

Hence, $\angle BAC = 110^o$.

Updated on 10-Oct-2022 13:46:41