Chapter 8.3, Problem 47AYU

Major League Baseball Field A major league baseball diamond is actually a square 90 feet on a side. The pitching rubber is located $60.5$ feet from home plate on a line joining home plate and second base.

a. How far is it from the pitching rubber to first base?

b. How far is it from the pitching rubber to second base?

c. If a pitcher faces home plate, through what angle does he need to turn to face first base?

To determine

To find:

a. How far is it from the pitching rubber to first base?

b. How far is it from the pitching rubber to second base?

c. The pitcher needs to turn at an angle of $92.8°$

Answer to Problem 47AYU

Solution:

a. The pitching rubber is $63.7$ feet from the 1^st base.

b. The pitching rubber is at a distance of $66.78$ feet from the second base

c. The pitcher needs to turn at an angle of $92.8°$

Explanation of Solution

Given:

A major league baseball diamond is actually a square 90 feet on a side. The pitching rubber is located $60.5$ feet from home plate on a line joining home plate and second base.

Formula used:

$c^2 = a^2 + b^2 - 2ab\text{cos}C$

Calculation:

a. To find how far is it from the pitching rubber to first base, we need to find x.

$a = 60.5, b = 90 \text{and} c = x, C = 45°$

$x^2 = {60.5}^2 + {90}^2 - 2 * 60.5 * 90 * \text{cos}45$

$x^2 = {55}^2 + {330}^2 - 2 * 55 * 330\text{cos}10°$

$x \approx 63.7$

The pitching rubber is $63.7$ feet from the 1^st base.

b. To find how far is it from the pitching rubber to second base, we need to find y.

$a = 90, b = 90 \text{and}, C = 90°$

$c^2 = {90}^2 + {90}^2$

$c = 90\surd 2 = 127.28$

$y = 127.28 - 60.5$

$y = 66.78$

The pitching rubber is at a distance of $66.78$ feet from the second base

c. we need to find angle $\text{A}$

$\text{cos}A = \frac{x^2 + y^2 - {90}^2}{2xy}$

$\text{cos}A = \frac{{60.5}^2 + {\mathrm{63..7}}^2 - {90}^2}{2 * 63.7 * 60.5}$

$A = 92.8°$

The pitcher needs to turn at an angle of $92.8°$