Question

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

• A. ${110}^o$
• B. ${115}^o$
• C. ${100}^o$
• D. ${120}^o$

Answer 1

The correct option is A ${110}^o$

$\angle CED+\angle BEC={180}^o$ [Linear pair]
$\Rightarrow \angle CED+{130}^o={180}^o$
$\Rightarrow \angle CED+{180}^o-{130}^o={50}^o$ ....(i)
$\angle ECD={20}^o$
In $\Delta CED$,
$\angle CED+\angle ECD+\angle CDE={180}^o$
[Sum of all the angles of a triangle is ${180}^o$]
$\Rightarrow {50}^o+{20}^o+\angle CDE={180}^o$ [Using (i)and (ii)]
$\Rightarrow {70}^o+\angle CDE={180}^o$
$\Rightarrow \angle CDE={180}^o-{70}^o$
$\Rightarrow \angle CDE={110}^o$ ...(iii)
Now $\angle BAC=\angle CDE={110}^o$
[Angle in the same segment of a circle are equal]