Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

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

Now, ray OP stands on the line TOQ.

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

Similarly, ray OS stands on the 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, $\frac{(\angle POQ+\angle QOR+\angle SOR+\angle POS)}{3}$ $=$ $\frac{{360}^{\circ }}{3}$ $=$ ${120}^{\circ }$