Asked by Nikki

A baseball diamond is a square with sides 22.4m The pitchers mound is 16.8m from home plate on the line joining home plate and second base. How far is the pitchers mound from first base?
Answered by Steve
see related questions below
Answered by Nikki
They aren't explained in a way that I can understand them. Can you explain more clearly?
Answered by Steve
Don't know. I supplied both solutions, one using the pythagorean theorem, and one using the law of cosines.

pick one, and tell me the first place you get stuck. I'll try to walk you through it. I prefer the law of cosines, since it's for a trig class. If you don't know that, read about it and come on back.
Answered by Nikki
H=Home
P=Pitcher
S=Second Base
HP=16.8
HF=22.4
Find distance PF

d^2=16.8^2+22.4^2-2(16.8)(22.4)cos H
d^2=282.24+501.76-752.64cos
d^2=31.36cosH
Stuck from here....
Answered by Steve
Let F be first base.

You know the angle PHF is 45 degrees, because the line to the mound is on the diagonal of the square.

So, cosH is easy to figure. Multiply, take square root, and you have d.

d = PF
Answered by Nikki
okay so,
d^2=282.24+501.76-752.64(cos45°)
=22.1749sqrt
=4.7090
Not the same answer you got.
Answered by Steve
So, what did you get for cos 45 degrees?
When you evaluate 282.24+501.76-752.64(cos45°) what do you get?

I get 251.8, not 22.1749
Answered by Nikki
Steve, you are the best! Thanks for your patience.
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