# Can some verify this problem?

1. Oct 23, 2007

### Flappy

1. The problem statement, all variables and given/known data
The problem has to do with laws of cosines/sines.

"The baseball player in center field is playing approximately 330 feet from the television camera that is behind home plate. A batter hits a fly ball that goes to the wall 420 feet from the camera. The camera turns 6 degrees to follow the play. Approximate the distance the center field has to run to make the catch."

The diagram given looks roughly like this:

2. Relevant equations

There are 3 equations relating to the Laws of Cosines:

1. a^2 = b^2 + c^2 - 2bccos(A)
2. b^2 = a^2 + c^2 - 2accos(B)
3. c^2 = a^2 + b^2 - 2abcos(C)

3. The attempt at a solution

At this point the case of this problem is SAS (side side angle). So I know that laws of cosines has to be used here.

I then labeled the diagram like this:

Givens:
b=420
c=330
A=6 degrees
a = ?

a is what needs to be solved.

I then used equation one since it has the most related givens

a^2 = b^2 + c^2 - 2bccos(A)
a^2 = 420^2 + 330^2 - 2(420)(330)cos6

a = $$\sqrt{420^2 + 330^2 - 2(420)(330)cos6}$$

a = 98.07ft

Thats my approximate answer. Can someone verify it?

2. Oct 23, 2007

### EnumaElish

The expression in the root is correct.

3. Oct 28, 2007

### Flappy

So is everything correct?

4. Oct 28, 2007

### pooface

Since you are taking the root of a number it should be +-. But negative distance doesn't make sense, but it is something to keep in mind.