Trigonometry word problem involving wheel? A wheel 5 feet in diameter rolls up with an incline of 18 degrees 20 minutes. What is the height of the center of the wheel above the base of the incline when the wheel has rolled 5 ft up the incline?
 A: Does this diagram help? It is not drawn to scale but it can help in visualizing the problem.

So note that $|BD|=5ft$, $|CD|=2.5ft$, $\angle BDC=90^\circ$. Having settled that $$|CM|=\dfrac{2.5}{\sin 72}=2.628655561\ldots$$ we now need to find $$|MD|=\dfrac{2.5}{\tan 72}=.812229924\ldots$$ Since $|BD|=|BM|+|MD|$, then $$|BM|=5-|MD|=4.187700759\ldots$$ Calculating now $$|ME|=|BM|\sin 18=1.294070702\ldots$$ Finally, $$|CE|=|CM|+|ME|.$$
A: Place an additional label where the segment joining $C$ to $E$ and the segment joining $B$ to $D$ intersect.  Let's agree to call that point $M$.  The length of $\overline{EM}$ is $|\overline{BM}|\sin(18^\circ)$.  The angle $\angle CMD$ is $72^\circ$.  Can you figure the rest?
Now use the fact that 
$$\sin(72^\circ) = {2.5\over |\overline{CM}|}.$$
so $$|\overline{CM}| = {2.5\over \sin(72^\circ)} = 2.629. $$
Also we have $|\overline{BM}| = 5 - | \overline{MD}|$.  Can you find the length of $\overline
{ MD}$?
A: Rotation of axes. Plug in
$$ x = 5^{'}, y =2.5 ^{'}, \alpha = 18 \frac 13^{0}$$
into 
$$ y_2 = x  \sin \alpha + y  \cos \alpha. $$
A: You'll just have to imagine that the radius (Half of the diameter which is 2.5 feet) is aligned with the height of the triangle. So You just have to get the height of the triangle. To get it's height. Simply...
Sin 18.33 = height / 5 feet
Sin 18.33 x 5 feet = height
(0.3145)(5 feet) = height
1.5725 = Height
So, to get the height of the center of the circle to the ground. Simply...
Height of the center of the circle  w/respect to the base = 1.5725 + 2.5
The answer would be...
4.0725 feet
