Chapter 8.3, Problem 48AYU

Little League Baseball Field According to Little League baseball official regulations, the diamond is a square 60 feet on a side. The pitching rubber is located 46 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?

Answer to Problem 48AYU

a. The pitching rubber is $42.58$ feet from the ${\text{1}}^{\text{st}}$ base.

Explanation of Solution

Given:

According to Little League baseball official regulations, the diamond is a square 60 feet on a side. The pitching rubber is located 46 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 $x$.

$a=46,b=60$ and $c=x,C={45}^o$

$x^2={46}^2+{60}^2-2*46*60*cos45$

$x²=5716+3600-3903$

$x\approx 42.58$ feet.

The pitching rubber is $42.58$ feet from the $1^{\text{st}}$ base.

To determine

To find:

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

Answer to Problem 48AYU

b. The pitching rubber is at a distance of $38.85$ feet from the second base.

Explanation of Solution

Given:

According to Little League baseball official regulations, the diamond is a square 60 feet on a side. The pitching rubber is located 46 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:

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

$a=60,b=60$ and $C=90°$

$c^2={60}^2+{60}^2=\left(46+y\right)²$

$c=60\sqrt{2}\approx 84.85$

$y=84.85-46$

$y=38.85$ feet.

The pitching rubber is at a distance of $38.85$ feet from the second base.

To determine

To find:

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

Answer to Problem 48AYU

c. The pitcher needs to turn at an angle of $85.2°$.

Explanation of Solution

Given:

According to Little League baseball official regulations, the diamond is a square 60 feet on a side. The pitching rubber is located 46 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:

c. we need to find angle $A$.

$\text{cos}A=\frac{x^2+y^2-{60}^2}{2xy}$

$\text{cos}A=\frac{{46}^2+{42.58}^2-{60}^2}{2*46*42.58}$

$A\approx 85.2°$

The pitcher needs to turn at an angle of $85.2°$.