Question

Question 5
In the given 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 $\angle BAC$.

Answer 1

Given,
$\angle BEC={130}^{\circ }$
$\angle ECD={20}^{\circ }$
$\angle BEC+\angle DEC={180}^{\circ }$ [Linear pair]
${130}^{\circ }+\angle DEC={180}^{\circ }$
$\angle DEC={180}^{\circ }-{130}^{\circ }={50}^{\circ }$

Now, in $\Delta DEC$
$\angle DEC+\angle ECD+\angle CDE={180}^{\circ }$ [Angle sum property of a triangle]
${50}^{\circ }+{20}^{\circ }+\angle CDE={180}^{\circ }$
$\angle CDE={180}^{\circ }-{70}^{\circ }={110}^{\circ }$
$\angle CDE=\angle BAC$ [Angles in the same segment]
$\therefore \angle BAC={110}^{\circ }$