# Thread: Word problem has to do with law of cosines

1. ## Word problem has to do with law of cosines

In major league baseball, the four bases form a square whose sides are each 90 feet long. The front edge of the pitching rubber on which the pitcher stands is 60.5 feet from home plate.

a) Find the distance from the center of the front edge of the pitching rubber to first base.
b) Find the distance from the center of the front edge of the pitching rubber to second base.
c) Find the angle created by home plate, the center of the front edge of the pitching rubber, and first base.
d) Find the angle formed by the center of the front edge of the pitching rubber, first base, and home plate.

Thanks a lot!

2. Hello, gobbajeezalus!

In major league baseball, the four bases form a square with 90-foot sides.
The pitcher's mound is 60.5 feet from home plate.
. . Assume the pitcher's mound is a point.
Code:
                C
*
*   *
*       *
90 *           * 90
*               *
*                   *
D *           P           * B
*         *         *
*       |       *
90 * 60.5|     * 90
*   |45°*
* | *
*
A
Draw line segment $PB.$

a) Find the distance from the pitcher's mound to first base.
In $\Delta PAB$, use the Law of Cosines:

$PB^2 \;=\;60.5^2 + 90^2 - 2(60.5)(90)\cos45^o \;=\;4059.857153$

Therefore: . $PB \;\approx\;63.72$ feet.

b) Find the distance from the pitcher's mound to second base.

In isosceles right triangle $CBA,\;CA = 90\sqrt{2}$

Then: . $PC \;=\;CA - PA \;=\;90\sqrt{2} - 60.5 \;\approx\;66.78$ feet.

c) Find the angle created by home plate, the pitcher's mound, and first base.
In $\Delta BPA$, use the Law of Cosines . . .

$\cos \angle BPA \;=\;\frac{60.5^2 + 63.72^2 - 90^2}{2(60.5)(63.72)} \;=\;-0.049222528$

Hence: . $\angle BPA \;=\;92.82138321 \;\approx\;92.82^o$

d) Find the angle formed by the pitcher's mound, first base, and home plate.

$\angle PBA \;=\;180^o - 45^o - 92.82^o \;=\;42.18^o$