Tuesday

January 24, 2017
Posted by **Nikki** on Tuesday, October 16, 2012 at 6:14pm.

- Trig -
**Steve**, Tuesday, October 16, 2012 at 6:16pmsee related questions below

- Trig -
**Nikki**, Tuesday, October 16, 2012 at 6:21pmThey aren't explained in a way that I can understand them. Can you explain more clearly?

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**Steve**, Tuesday, October 16, 2012 at 6:35pmDon'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. - Trig -
**Nikki**, Tuesday, October 16, 2012 at 7:20pmH=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.... - Trig -
**Steve**, Tuesday, October 16, 2012 at 9:36pmLet 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 - Trig -
**Nikki**, Tuesday, October 16, 2012 at 9:56pmokay so,

d^2=282.24+501.76-752.64(cos45°)

=22.1749sqrt

=4.7090

Not the same answer you got. - Trig -
**Steve**, Wednesday, October 17, 2012 at 10:19amSo, 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 - Trig -
**Nikki**, Wednesday, October 17, 2012 at 3:37pmSteve, you are the best! Thanks for your patience.