Question

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

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

Answer 1

Answer: (C)

${90}^o$

Let, $\angle QSR+\frac{P}{2}=x$ ..................(1)

Now, in a $△QRS$,

$\angle RQS+\angle QRS+\angle QSR={180}^{\circ }$ .................. (2)

From (1) and (2),

${180}^{\circ }-\angle RQS-\angle QRS+\frac{\angle QPR}{2}=x$ .................. (3)

In a $△PQR$,

$\angle QPR={180}^{\circ }-\angle QRP-\angle PQR$

Divide the above equation by $2$, we get

$\frac{\angle QPR}{2}={90}^{\circ }-\frac{\angle QRP}{2}-\frac{\angle PQR}{2}$ .................. (4)

From (3) and (4), we get

${270}^{\circ }-\angle RQS-\angle QRS-\frac{\angle QRP}{2}-\frac{\angle PQR}{2}=x$ .................. (5)

Now, $\angle PQR+2\angle RQS={180}^{\circ }$

Dividing by $2$ , we get

$\frac{\angle PQR}{2}+\angle RQS={90}^{\circ }$ .................. (6) (Linear pair and exterior bisector)

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

Dividing by 2, we get

$\frac{\angle QRP}{2}+\angle QRS={90}^{\circ }$ .................. (7) (Linear pair and exterior bisector)

From (5),(6) and (7), we get

${270}^{\circ }-{90}^{\circ }-{90}^{\circ }=x$

$x={90}^{\circ }$

Hence, option C is the correct answer.