Question

In a $\Delta PQR$, the sides $PQ$ and $PR$ are produces to $S$ and $T$ respectively. Bisectors of $\angle SQR$ and $\angle QRT$ meet at the point $O$. If $\angle P={66}^{\circ }$, then what is the value of $\angle QOR$?

• A. ${47}^{\circ }$
• B. ${50}^{\circ }$
• C. ${57}^{\circ }$
• D. ${67}^{\circ }$

Answer 1

The correct option is C ${57}^{\circ }$

(c) In $\Delta PQR$,
$\angle PQR+\angle PRQ+\angle QPR={180}^{\circ }$
$\Rightarrow \angle PQR+\angle PRQ={180}^{\circ }-\angle QPR$
$={180}^{\circ }-{66}^{\circ }$
$={114}^{\circ }$
Now, $\angle SQR={180}^{\circ }-\angle PQR...\left(i\right)$
and $\angle QRT={180}^{\circ }-\angle PRQ....\left(ii\right)$
Adding eq. $\left(i\right)$ and $\left(ii\right)$, we get
$\Rightarrow \angle SQR+\angle QRT={360}^{\circ }-(\angle PQR+\angle PRQ)$
In $\Delta QOR$,
$\angle RQO+\angle QRO+\angle QOR={180}^{\circ }$
$\Rightarrow \angle QOR={180}^{\circ }-(\angle RQO+\angle QRO)$
$={180}^{\circ }-\left(\begin{array}{c}\frac{1}{2}\angle SQR+\frac{1}{2}\angle QRT\end{array}\right)(\because QObisects\angle SQR,RObisects\angle QRT)$
$={180}^{\circ }-\frac{1}{2}(\angle SQR+\angle QRT)$
$={180}^{\circ }-\frac{1}{2}[{360}^{\circ }-(\angle PQR+\angle PRQ\left)\right]$
$={180}^{\circ }-{180}^{\circ }+\frac{1}{2}(\angle PQR+\angle PRQ)$
$=\frac{1}{2}\times {114}^{\circ }={57}^{\circ }$.