posted by Kailey .
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)
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