Question

In this Figure, $QS$ and $RS$ are the bisectors of exterior angles $Q$ and $R$ respectively. Then $\angle QSR$ $+$ $\frac{\angle P}{2}$ is equal to:

• A. ${270}^{\circ }$
• B. ${180}^{\circ }$
• C. ${90}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is C ${90}^{\circ }$

$\angle QSR+\frac{P}{2}=x$....(1)
Now, in a triangle $QRS$,
$\angle RQS+\angle QRS+\angle QSR={180}^o$....(2)
From (1) and (2),
${180}^o-\angle RQS-\angle QRS+\angle QPR/2=x$....(3)
In a triangle $PQR$,
$\angle QPR=180-\angle QRP-\angle PQR$
Divide the above equation by 2, we get
$\angle QPR/2=90-\angle QRP/2-\angle PQR/2$....(4)
From (3) and (4), we get
${270}^o-\angle RQS-\angle QRS-\angle QRP/2-\angle PQR/2=x$.....(5)
Now, $\angle PQR+2\angle RQS=180$
Dividing by 2 , we get
$\angle PQR/2+\angle RQS=90$......(6) (Linear pair and exterior bisector)
$\angle QRP+2\angle QRS=180$
Dividing by 2, we get
$\angle QRP/2+\angle QRS=90$.......(7) (Linear pair and exterior bisector)
From (5),(6) and (7), we get
${270}^o-{90}^o-{90}^o=x$
$x={90}^o$
Hence, option $C$ is the correct answer.