Question

Two circles intersect at $\text{B}$ and $\text{E}$. Quadrilaterals $\text{ABEF}$ and $\text{BCDE}$ are inscribed in these circles such that $\text{ABC}$ and $\text{FED}$ are line segments. Also, $\angle \text{A}={96}^{\circ }$ and $\angle \text{F}={75}^{\circ }$, the value of $x$ is:

• A. ${96}^{\circ }$
• B. ${84}^{\circ }$
• C. ${75}^{\circ }$
• D. ${105}^{\circ }$

Answer 1

The correct option is B ${84}^{\circ }$

Given that,
$\angle \text{A}={96}^{\circ }$
$\angle \text{F}={75}^{\circ }$
In the quadrilateral $\text{ABEF}$,
$\angle \text{A}+\angle \text{BEF}={180}^{\circ }$ ( In cyclic quadrilateral, sum of opposite angles are supplementary)
$\angle \text{BEF}={180}^{\circ }-{96}^{\circ }$
$\angle \text{BEF}={84}^{\circ }$

Now, $\angle \text{BEF}+\angle \text{BED}={180}^{\circ }$
$\Rightarrow \angle \text{BED}={180}^{\circ }-{84}^{\circ }$
$={96}^{\circ }$

In the quadrilateral $\text{BCDE}$,
$\angle \text{BED}+\angle \text{C}={180}^{\circ }$
$\because \angle \text{BED}={96}^{\circ }$
$\angle \text{C}+{96}^{\circ }={180}^{\circ }$
$\angle \text{C}={180}^{\circ }-{96}^{\circ }$
$\Rightarrow \angle \text{C}={84}^{\circ }$
$\Rightarrow x={84}^{\circ }$
So, $x={84}^{\circ }$