Question

In the figure below, OP, OQ, OR and OS are four rays. Then, find $\frac{(\angle POQ+\angle QOR+\angle SOR+\angle POS)}{3}$.

• A. ${60}^{\circ }$
• B. ${120}^{\circ }$
• C. ${90}^{\circ }$
• D. $0^{\circ }$

Answer 1

The correct option is B

${120}^{\circ }$

Produce ray OQ backwards to a point T, so that, TOQ is a line.

Now, ray OP stands on line TOQ.

Therefore, $\angle$TOP + $\angle$POQ = ${180}^{\circ }$ ---- (1) (linear pair axiom)

Similarly, ray OS stands on line TOQ.

Therefore, $\angle$TOS + $\angle$SOQ = ${180}^{\circ }$ ----(2)

But, $\angle$SOQ = $\angle$SOR + $\angle$QOR

So, (2) becomes,
$\angle$TOS + $\angle$SOR + $\angle$QOR = ${180}^{\circ }$ ----(3)

Now, adding (1) and (3), we get,

$\angle$TOP + $\angle$POQ + $\angle$TOS + $\angle$SOR + $\angle$QOR = ${360}^{\circ }$ ----(4)

But, $\angle$TOP + $\angle$TOS = $\angle$POS

Hence, (4) becomes,
$\angle$POQ + $\angle$QOR + $\angle$SOR + $\angle$POS = ${360}^{\circ }$

So, $\frac{(\angle POQ+\angle QOR+\angle SOR+\angle POS)}{3}$ = $\frac{{360}^{\circ }}{3}$ = ${120}^{\circ }$

We also know that angles around a point = ${360}^{\circ }$

Hence, $\frac{{360}^{\circ }}{3}$ = ${120}^{\circ }$