Chapter 8.3, Problem 47AYU

a.

To determine

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

Answer to Problem 47AYU

  $\boxed{63.7}$ ft

Explanation of Solution

Given information:

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

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

Calculation:

The base ball field under consideration will be shown in the figure.

  

Let x be the distance from the pitching rubber to the first base. The two sides and the included of the triangle in the figure are known. So, we can use the Law of Cosines to find x. $x^2={60.5}^2+{90}^2-\mathrm{2.60.5.90.}\cos 45°$

  $=3660.25+8100-7700.39$

  $=4059.86$

Take square root on both the sides.

  $\sqrt{x^2}=\sqrt{4059.86}$

  $x=63.7$

Hence, the pitching rubber is 63.7ft far from the first base.

b.

To determine

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

Answer to Problem 47AYU

  $\boxed{66.8ft}$

Explanation of Solution

Given information:

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

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

Calculation:

Since the field is a square of side 90ft, the distance from the home to the second base is equivalent to the diagonal of the square.

In the figure we can see that the diagonal $y+\mathrm{60.5.}$

The diagonal of a square of side length a is $\sqrt{2a.}$

  $\sqrt{2.90}=y+60.5$

  $127.3=y+60.5$

Subtract 60.5 from both the sides.

  $127.3-60.5=y+60.5-60.5$

  $66.8=y$

Hence, the pitching rubber is 66.8 ft far from the second base.

c.

To determine

What angle does he need to turn to face first base?

Answer to Problem 47AYU

  $\boxed{92.8°}$

Explanation of Solution

Given information:

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

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

Calculation:

Let the pitcher has to turn an angle of $\theta$ to face the first base.

We know the three sides of the triangle. So, using the Law of Cosines, we can find $\theta$ .

  $\begin{array}{l}{90}^2={60.5}^2+{63.7}^2-\mathrm{2.60.5.63.7.}\cos \theta \\ 8100=3660.25+4057.69-7707.7\cos \theta \end{array}$

  $\theta ={\cos }^{-1}\left(\frac{3660.25+4057.69-8100}{7707.7}\right)$

  $\theta ={\cos }^{-1}\left(-0.049\right)$

  $\approx 92.8°$

Hence, the pitcher has to turn through an angle of about $92.8°$ to face the first base.