2013-I-06

Since $L$ is the angle bisector of $\mathrm{\angle }AOB$, then the angle between $L$ and $OA$ is $({130}^{\circ }-{10}^{\circ })÷2={60}^{\circ }$.

Note that $\mathrm{\Delta }OAB$ is an isosceles triangle with $OA=OB$, then $L\perp AB$. Hence, we have

$\begin{array}{rcl}\cos {60}^{\circ }& =& {\displaystyle \frac{r}{OA}}\\ r& =& 26\times \cos {60}^{\circ }\\ & =& 13\end{array}$

Also,

$\begin{array}{rcl}\theta & =& {10}^{\circ }+{60}^{\circ }\\ & =& {70}^{\circ }\end{array}$

Therefore, the polar coordinates of the intersection point are $(13,{70}^{\circ })$.