Question

Two circles intersect in A and B. Quadrilaterals PCBA and ABDE are inscribed in these circles such that PAE and CBD are line segments. If $\angle P={95}^{\circ }and\angle C={40}^{\circ }$. Find the value of $\angle DEA$.

• A. ${65}^{\circ }$
• B. ${105}^{\circ }$
• C. ${95}^{\circ }$
• D. ${85}^{\circ }$
  

Answer 1

The correct option is D ${85}^{\circ }$


In $\Delta ABC,$
$\angle P+\angle CBA={180}^{\circ }$ [Opposite angles of a quadrilateral add upto 180${}^{\circ }$]
$\Rightarrow \angle CBA={180}^{\circ }-{95}^{\circ }={85}^{\circ }$
$\angle CBD={180}^{\circ }$ [Angle of a straight line]
$\Rightarrow \angle ABD={180}^{\circ }-\angle CBD={180}^{\circ }-{85}^{\circ }={95}^{\circ }$ [$\angle ABD$ and $\angle CBD$ are supplementary to each other]
In $ABDE,\angle B+\angle E={180}^{\circ }$ [Opposite angles of a quadrilateral add upto 180${}^{\circ }$]
Let $\angle E=z^{\circ }={180}^{\circ }-\angle ABD={180}^{\circ }-{95}^{\circ }$ $={85}^{\circ }$