Chapter 12.4, Problem 4GP

To determine

To find: How far away from third base was the ball caught.

Answer to Problem 4GP

The distance is about $85.74$ feet.

Explanation of Solution

Given information:

  $\begin{array}{l}A={43}^0\\ C={65}^0\\ b=90ft\end{array}$

Let Point A be Third base and Point C be Second base

B is point at which the ball caught.

Concept:

Law of sines

  $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

Given

  $\begin{array}{l}A={43}^0\\ C={65}^0\\ b=90ft\end{array}$

We know sum of Angles of Triangle is ${180}^0$

Using that

We get

  $\begin{array}{l}A+B+C={180}^0\\ \Rightarrow {43}^0+B+{65}^0={180}^0\\ \Rightarrow B+{108}^0={180}^0\\ \Rightarrow B={180}^0-{108}^0\\ \Rightarrow B={72}^0\end{array}$

Using law of sines

We get

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\\ \Rightarrow \frac{\sin {72}^0}{90}=\frac{\sin \left({65}^0\right)}{c}\\ \Rightarrow c=\frac{90\sin \left({65}^0\right)}{\sin \left({72}^0\right)}\\ \Rightarrow c=\frac{90\left(0.906\right)}{\left(0.951\right)}\\ \Rightarrow c=\frac{81.54}{\left(0.951\right)}\\ \Rightarrow c=85.74\end{array}$

So, the distance is about $85.74$ feet.