# 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?

-
Draw a picture. Do you see a right triangle that might give you some clues. – ncmathsadist Jul 3 '12 at 21:39
I drew a picture with 5 being hypotnouse 18 degrees 20 being an angle and I got 1.57. When i did sin 18 degrees 20= x/5. – Fernando Martinez Jul 3 '12 at 22:01
Hmm I am not sure how to proceed. – Fernando Martinez Jul 3 '12 at 22:10
@Bishop did you figure it out? – azetina Jul 3 '12 at 22:11
No I cant seem to figure it out. – Fernando Martinez Jul 3 '12 at 22:11

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, $$|CE|=|CM|+|ME|.$$

-
Similar trangles galore! – ncmathsadist Jul 3 '12 at 22:17
Yes I did something similar and I got BE is 4.74 when I did 5 cos 18 degrees 20 minutes the thing is my final answer is 3.75 ft and I am not sure how to get there. I mean I know from E to right below the circle is 1.57. – Fernando Martinez Jul 3 '12 at 22:18
Maybe I add 1.57 plus the radius which is 2.5 feet. – Fernando Martinez Jul 3 '12 at 22:20
@azetina Good Drawing! – ncmathsadist Jul 3 '12 at 22:22
@ncmathsadist thanks. I did it with GeoGebra. Very nice program indeed. – azetina Jul 3 '12 at 22:23

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 $\overline { MD}$?

-
I know the lenght of EM is about 1.57 but no I am not sure about figuring out the rest. Am I to use 72 degrees to find a certain length? – Fernando Martinez Jul 3 '12 at 22:24
@ncmathsadist I disagree with you since EM is in $\triangle BEM$ and therefore its hyp0tenuse is not 5 ft. – azetina Jul 3 '12 at 22:30
I will say the final answer is 3.95 I know BE is 4.74 the question is how to reach the final answer? – Fernando Martinez Jul 3 '12 at 22:32
would MD be .81240 I did 2.629 cos 72 – Fernando Martinez Jul 3 '12 at 22:43

Rotation of axes. Plug in

$$x = 5^{'}, y =2.5 ^{'}, \alpha = 18 \frac 13^{0}$$

into $$y_2 = x \sin \alpha + y \cos \alpha.$$

-

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

-