1. ## Strange trigonometry question...

Not sure on how to set this one up, some sort of strange triangle.
Staring down the pitcher on a full count, Phil, whose eyes are at 5'4", notices that he looks down 4° to see the pitching rubber 60' away. How tall is the mound in inches?
Can anyone help me set this strange baseball word problem up?

2. Originally Posted by OmegaThetaOmega
Not sure on how to set this one up, some sort of strange triangle. Can anyone help me set this strange baseball word problem up?
We have a right triangle with a horizontal leg of 60' = 720" and a vertical leg of 5'4" + h = 64" + h, calling h the height of the mound.

We know that the angle between the 720" leg and the hypotenuse is 4 degrees.

Thus we have the relation:
$\displaystyle tan(4^o) = \frac{64 + h}{720}$

Can you finish from here?

-Dan

3. I got $\displaystyle h \approx -13.7$. I would accept this answer, but distance can't be negative.

4. Originally Posted by OmegaThetaOmega
I got $\displaystyle h \approx -13.7$. I would accept this answer, but distance can't be negative.
You are correct. Apparently the problem sets up an impossible condition. (I hadn't solved for h myself until now so I didn't spot this. Sorry!)

-Dan