Question

In the given figure, $AB=CD$ and $\overline{AB}$ and $\overline{CD}$ are produced to meet at the point $P$.
If $\angle BAC={70}^0$, then $\angle P$ is

• A. ${30}^0$
• B. ${40}^0$
• C. ${45}^0$
• D. ${50}^0$

Answer 1

Answer: (B)

${40}^0$

Exterior angle of a cyclic quadrilateral is equal to its interior opposite angle.
$\Rightarrow \angle BAC=\angle DCA={70}^0$
In $\Delta PAC,{70}^0+{70}^0+\angle P={180}^0$
$\Rightarrow \angle P={40}^0$