Answer
The two roads measure $404\text{ and }551\text{ feet}$ respectively.
Work Step by Step
For the given triangle ABC, $A=55{}^\circ,B=46{}^\circ,a=460$
Using the angle sum property, compute the measure of angle C as follows:
$\begin{align}
& 55{}^\circ +46{}^\circ +C=180{}^\circ \\
& C=180{}^\circ -101{}^\circ \\
& C=79{}^\circ
\end{align}$
Now, using the law of sines we will obtain the value of both b and c:
$\begin{align}
& \frac{a}{\sin A}=\frac{c}{\sin C} \\
& \frac{460}{\sin 55{}^\circ }=\frac{c}{\sin 79{}^\circ } \\
& c\cdot \sin 55{}^\circ =460\cdot \sin 79{}^\circ \\
& c=\frac{460\cdot \sin 79{}^\circ }{\sin 55{}^\circ }
\end{align}$
This gives $c\approx 551$ to the nearest tenth.
Also,
$\begin{align}
& \frac{a}{\sin A}=\frac{b}{\sin B} \\
& \frac{460}{\sin 55{}^\circ }=\frac{b}{\sin 46{}^\circ } \\
& b\cdot \sin 55{}^\circ =460\cdot \sin 46{}^\circ \\
& b=\frac{460\cdot \sin 46{}^\circ }{\sin 55{}^\circ }
\end{align}$
This gives $b\approx 404$ to the nearest tenth.
