Posted by **Kailey** on Monday, October 1, 2012 at 11:03am.

A baseball diamond is a square sides 22.4 m. The pitcher's mound is 16.8 m from home plate on the line joining home plate and second base.. How far is the pitcher's mound from first base? (we are using the LAW OF SINES AND COSINES)

- Math -
**Steve**, Monday, October 1, 2012 at 11:16am
letting home be H, pitcher be P, 2nd base be S, we have triangle HSF where

HP = 16.8

HF = SF = 22.4

we want d = PF

d^2 = 16.8^2 + 22.4^2 - 2(16.8)(22.4)cosH

Now, the diagonal from HS to 2nd base is 22.4√2 = 31.7m

In triangle HSF,

22.4^2 = 31.7^2 + 22.4^2 - 2(31.7)(22.4)cosH

cosH = .7076

Note that H is not quite 45°, since P is not quite at the midpoint of HS.

So, now we have

d^2 = 16.8^2 + 22.4^2 - 2(16.8)(22.4)(.7076)

d = 15.85m

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