Basic trigonometry truck problem I've got a decent grasp on trig but this one problem annoys me so much I need to get it answered. Here it is:
"A truck with a long back tray can scrape the road when it travels up steep driveways. If the base of the tray is 1 metre off the road, the road slopes off the gutter at 10 degrees, and the driveway slopes off the gutter at 15 degrees, what is the longest back tray that doesn't scrape the road? Take the extension to be the measurement from above the back axle to the end of the truck."
I know the answer is 2.14 metres but I just can't find the process to get there. Thanks in advance!
 A: Consider the figure below:

The driveway and the road intersect at point $G$.  The cross-section of the roadway, represented by $\overline{GR}$ in the figure forms an angle of $10^\circ$ with the gutter, which is assumed to be horizontal and perpendicular to the plane of the figure.  The cross-section of the driveway, represented by $\overline{GD}$ in the figure forms an angle of $15^\circ$ with the gutter.  As the truck backs down the driveway, its tray, the side of which lies along $\overline{RT}$ in the figure, is parallel to the driveway.  When the back wheels of the truck meet the road, the back axle of the truck is at position $\overline{BG}$.  Since the truck's tray is one meter off the ground, $|BG| = 1~\text{m}$.    
As the figure indicates, $\overline{GV}$ is vertical.  Thus, $\angle VGD$ is complementary to $\angle DGF$.  Hence, $m\angle VGD = 90^\circ - 15^\circ = 75^\circ$.  If two parallel lines are cut by a transversal, then alternate interior angles are congruent, so $m\angle BVG = m\angle VGD = 75^\circ$.  Since $\angle VBG$ is a right angle, $\angle VGB$ is complementary to $\angle BVG$.  Hence, $m\angle VGB = 90^\circ - 75^\circ = 15^\circ$. By the Angle Addition Postulate, 
$$m\angle AGR + m\angle RGB + m\angle BGV = m\angle AGV$$
Since $\angle AGV$ is a right angle, 
$$m\angle AGR + m\angle RGB + m\angle BGV = 90^\circ$$
Substituting $10^\circ$ for $m\angle AGR$ and $15^\circ$ for $m\angle BGV$ and solving for $m\angle RGB$ yields 
\begin{align*}
10^\circ + m\angle RGB + 15^\circ & = 90^\circ\\
m\angle RGB & = 65^\circ
\end{align*}
If the back tray of the truck is long enough to just meet the road, then the length of the back tray is $b = |BR|$.  Since 
$$\tan\angle RGB = \frac{b}{1~\text{m}}$$
and $m\angle RGB = 65^\circ$, we obtain
$$b = (1~\text{m})\tan(65^\circ) \approx 2.14450692051~\text{m}$$
Hence, the longest tray that does not meet the road is, to the nearest centimeter, $2.14~\text{m}$.  That said, a tolerance of just $4.5~\text{mm}$ seems like a bad idea to me.  
