Question

In the figure below, lines RS and PQ intersect each other at point O. If $\angle$POR: $\angle$ROQ = 11:13, find $\angle$SOQ.

• A. $(\frac{165}{2}{)}^{\circ }$
• B. $(\frac{195}{2}{)}^{\circ }$
• C. ${95}^{\circ }$
• D. ${65}^{\circ }$

Answer 1

The correct option is A $(\frac{165}{2}{)}^{\circ }$

$\angle$POR + $\angle$ROQ = ${180}^{\circ }$ (linear pair of angles)

but, $\angle$POR : $\angle$ROQ = 11:13 (given)

$\angle$POR = $\frac{11}{24}$ $\times$ ${180}^{\circ }$ = $(\frac{165}{2}{)}^{\circ }$

Similarly, $\angle$ROQ = $\frac{13}{24}$ $\times$ ${180}^{\circ }$ = $(\frac{195}{2}{)}^{\circ }$

Now, $\angle$POS = $\angle$ROQ = $(\frac{195}{2}{)}^{\circ }$ (vertically opposite angle)

And, $\angle$SOQ = $\angle$POR= $(\frac{165}{2}{)}^{\circ }$ (vertically opposite angle)