Chapter 5.6, Problem 36E

Solution

a)

To find:The distance between point $B$ and far end of the second base.

The distance between point $B$ and far end of the second base is $66.779\text{ft}$ . This distance is $3\text{ft}$ longer as compared to the distance between pitcher’s rubber and the first base.

Given information:

The distance between point $A$ and point $C$ is $90\text{ft}$ . The distance between point $B$ and second base is $60.5\text{ft}$ . The included $\angle C=45°$ .

Formula used:

Law of Cosines:

  $c^2=a^2+b^2-2ab\cos C$

Calculation:

Consider the distance between point $B$ and far end of the first base as $c$ . Let the between point $A$ and point $C$ be $90\text{ft}$ and the distance between point $B$ and second base be $60.5\text{ft}$ .

  

Substitute $90\text{ft}$ for $AC$, $60.5\text{ft}$ for $BC$ and $45°$ for angle $C$ in $c^2=a^2+b^2-2ab\cos C$ to calculate the distance $c$ .

  $\begin{array}{c}{c}^2=a^2+b^2-2ab\cos C\\ c=\sqrt{a^2+b^2-2ab\cos C}\\ =\sqrt{{60}^2+{90}^2-2\times 60.5\times 90\cos 45°}\\ =63.717\end{array}$

Apply cosine function in $\Delta CAD$, $\cos \left(45°\right)=\frac{90}{CD}$ .

  $CD=127.279\text{ft}$

Distance between $C$ and $D$ is $BD$ .

  $\begin{array}{c}BD=127.279-60.5\\ =66.779\end{array}$

So, the distance between pitcher’s rubber and the first base as compared to point $A$ and point $C$ is $66.779-63.717=3\text{ft}$ .

Therefore, distance between point $B$ and far end of the second base is $66.779\text{ft}$ and this distance is $3\text{ft}$ longer as compared to the distance between pitcher’s rubber and the first base.

b)

To find:The measure of angle $B$ in $\Delta ABC$ .

The measure of angle $B$ in $\Delta ABC$ is $92.84°$ .

Given information:

The distance $AD$ is $63.717\text{ft}$, distance $BC$ is $60.5\text{ft}$, and distance $AC$ is $90\text{ft}$ . The included angle $\angle C=45°$ .

Formula used:

Law of Cosines:

  $b^2=a^2+c^2-2ac\cos B$

Calculation:

Consider the diagram.

  

Consider $\Delta ABC$ .

Substitute $90$ for $b$, $60.5$ for $a$, and $63.717$ for $c$ in $b^2=a^2+c^2-2ac\cos B$ to calculate the measure of $\angle B$ .

  $\begin{array}{c}{90}^2={60.5}^2+{63.717}^2-2\times 60.5\times 63.717\cos B\\ \angle B={\cos }^{-1}\left(-0.04927\right)\\ =92.824°\end{array}$

Therefore, the measure of $B$ in $\Delta ABC$ is $92.84°$ .