Question

Lines RS and PQ intersect each other at point O. If $\angle POR:\angle ROQ$ = 11:13, $\angle SOQ$ =

• A. ${82.5}^{\circ }$
• B. ${87.5}^{\circ }$
• C. ${92}^{\circ }$

Answer 1

The correct option is A ${82.5}^{\circ }$

$\angle POR+\angle ROQ={180}^{\circ }$ (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, $\angle SOQ=\angle POR=(\frac{165}{2}{)}^{\circ }={82.5}^{\circ }$ (vertically opposite angle)