Circle Geometry Problem

I was trying this problem and i couldnt come up with a solution, it doesnt seem to hard but I cant think of what to do.

There is a point on a wheel that is rotating at a constant speed. The bottom of the wheel is 1 meters above the ground. At a point Q the point on the wheel is 16 meters above the ground. After 4 seconds its reaches the top of the wheel. After another 8 seconds it reaches a point P. If the wheel has a diameter of 16 meters how do we find out how high the point is at point P.

Thanks!

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If you can use coordinates, they might help greatly. Center the rotating circle on the vertical axis; trigonometry might be helpful. –  Guess who it is. Apr 15 '12 at 16:03

To follow the argument below, you will need to draw a diagram.

Let $O$ be the centre of the wheel, and $T$ the top of the wheel.

Draw a perpendicular from $Q$ to the vertical line $OT$, meeting that line at $X$. The height of the point $Q$ (and therefore of $X$) is $16$, and the height of $T$ is $17$. It follows that $XT=1$ and therefore $OX=7$.

Let $\theta=\angle QOX$. We have $\cos\theta=\frac{7}{8}$. In $4$ seconds the wheel travels through an angle $\theta$. So in $8$ seconds it travels through an angle $2\theta$.

We can find $\cos 2\theta$ approximately, using a calculator, or exactly, using $\cos 2\theta=2\cos^2\theta-1$. Let's go for exactly. We get $\cos 2\theta=\frac{34}{64}$.

Now we can find the height of point $P$. The picture is much like the one we worked with for $Q$. Draw a perpendicular from $P$ to the vertical line $OT$, meeting that line at $Y$. Then $OY=8\cos 2\theta$. So the height of $P$ above the centre of the wheel is $8\cos 2\theta$, and therefore the height above the ground is $9+8\cos 2\theta$.

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