Question

In this figure QS and RS are bisectors of exterior angles Q and R. Then $\angle QSR+\angle P/2$ is equal to

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

Answer 1

The correct option is C ${90}^o$

In given $△$PAB,

QS, RS are the angle bisector of $\angle$AQR and $\angle$BRQ.

So, $\angle$AQS = $\angle$SQR and $\angle$QRS = $\angle$SRB --- (1)

Side PQ and PR of $△$PQR are produced to A and B respectively.

$\therefore$ Exterior of $\angle$AQR = $\angle$P + $\angle$R --- (2)

and Exterior of $\angle$BRQ = $\angle$P + $\angle$Q --- (3)

Adding (2) and (3) we het,

$\angle$AQR + $\angle$BRQ = $\angle$P + $\angle$R + $\angle$P + $\angle$Q

$\angle$AQR + $\angle$BRQ = 2$\angle$P + $\angle$R + $\angle$Q

2$\angle$SQR + 2$\angle$QRS = $\angle$P + 180${}^{\circ }$

$\angle$SQR + $\angle$QRS = $\frac{1}{2}$$\angle$P + 90${}^{\circ }$ --- (4)

But in $△$QSR,

$\angle$SQR + $\angle$QRS + $\angle$QSR = 180${}^{\circ }$ --- (5)

From equatiomn (4) and (5) we het,

$\frac{1}{2}\angle$P + 90${}^{\circ }$ + $\angle$QSR = 180${}^{\circ }$

$\therefore$ $\angle$QSR + $\frac{1}{2}\angle$ P = 90${}^{\circ }$