Question

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

• A. ${110}^{\circ }$
• B. ${130}^{\circ }$
• C. ${20}^{\circ }$
• D. ${150}^{\circ }$

Answer 1

The correct option is A ${110}^{\circ }$


Consider the $△\text{CDE}$,
$\angle \text{BEC}$ is exterior angle,
So,
$\angle \text{BEC}=\angle \text{EDC}+\angle \text{ECD}$
$\angle \text{EDC}=\angle \text{BEC}-\angle \text{EDC}$
$={130}^{\circ }-{20}^{\circ }$
$={110}^{\circ }$
$\Rightarrow \angle \text{EDC}=\angle \text{BDC}={110}^{\circ }$

However, $\angle \text{BAC}=\angle \text{BDC}$, (Angles in the same segment of a circle)
$\Rightarrow \angle \text{BAC}={110}^{\circ }$