Ans: (a) $L$ is the angle bisector of $\angle AOB$. (b) $(13,70^\circ)$
- $L$ is the angle bisector of $\angle AOB$.
- Let $(r,\theta)$ be the polar coordinates of the intersection point.
Since $L$ is the angle bisector of $\angle AOB$, then the angle between $L$ and $OA$ is $(130^\circ-10^\circ)\div 2 = 60^\circ$.
Note that $\Delta OAB$ is an isosceles triangle with $OA=OB$, then $L\perp AB$. Hence, we have
$\begin{array}{rcl}
\cos 60^\circ & = & \dfrac{r}{OA} \\
r & = & 26 \times \cos 60^\circ \\
& = &13
\end{array}$Also,
$\begin{array}{rcl}
\theta & = & 10^\circ+60^\circ \\
& = & 70^\circ
\end{array}$Therefore, the polar coordinates of the intersection point are $(13, 70^\circ)$.
