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)