In the figure,$OP$,$OQ$,$OR$,and $OS$ are four rays. Prove that $\angle POQ + \angle QOR + \angle SOR + \angle POS = 360^o$.

(N/A) To prove this,we need to produce any of the rays $OP$,$OQ$,$OR$,or $OS$ backwards to a point. Let us produce ray $OQ$ backwards to a point $T$ such that $TOQ$ is a straight line.
Now,ray $OP$ stands on line $TOQ$.
Therefore,$\angle TOP + \angle POQ = 180^o$ ........ $(1)$ (Linear pair axiom)
Similarly,ray $OS$ stands on line $TOQ$.
Therefore,$\angle TOS + \angle SOQ = 180^o$ ........ $(2)$
But,$\angle SOQ = \angle SOR + \angle QOR$.
Substituting this into $(2)$,we get:
$\angle TOS + \angle SOR + \angle QOR = 180^o$ ........ $(3)$
Now,adding $(1)$ and $(3)$,we get:
$\angle TOP + \angle POQ + \angle TOS + \angle SOR + \angle QOR = 360^o$ ........ $(4)$
Since $\angle TOP + \angle TOS = \angle POS$,equation $(4)$ becomes:
$\angle POQ + \angle QOR + \angle SOR + \angle POS = 360^o$.