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, $\angle BAC = 110^o$.

Tutorialspoint

Updated on: 10-Oct-2022

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