Question

The internal bisectors of $\angle Q$ and $\angle R$ of $△PQR$ meet at $C$. If $\angle P={70}^o$, then $\angle P={70}^o$ is

• A. ${110}^o$
• B. ${125}^o$
• C. ${130}^o$
• D. ${140}^o$

Answer 1

The correct option is B ${125}^o$

In $△PQR$

$Y+Z+70=180$(sum of angles)

$\Rightarrow Y+Z=110---1$

In $△QCR$

$\frac{Y}{2}+\frac{Z}{2}+\angle QCR={180}^0$

$\Rightarrow 55+\angle QCR={180}^0$

$\Rightarrow \angle QCR={125}^0$