# Strange trigonometry question...

• Sep 30th 2007, 08:17 PM
OmegaThetaOmega
Strange trigonometry question...
Not sure on how to set this one up, some sort of strange triangle. :confused:
Quote:

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? :)
• Sep 30th 2007, 08:32 PM
topsquark
Quote:

Originally Posted by OmegaThetaOmega
Not sure on how to set this one up, some sort of strange triangle. :confused: 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:
$tan(4^o) = \frac{64 + h}{720}$

Can you finish from here?

-Dan
• Sep 30th 2007, 08:40 PM
OmegaThetaOmega
I got $h \approx -13.7$. I would accept this answer, but distance can't be negative. :confused:
• Sep 30th 2007, 11:33 PM
topsquark
Quote:

Originally Posted by OmegaThetaOmega
I got $h \approx -13.7$. I would accept this answer, but distance can't be negative. :confused:

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