Chapter 13.4, Problem 40PPS

a.

To determine

To find:Distance between Miami and Bermuda.

Answer to Problem 40PPS

Distance between sirens B and C is $15.8$ miles

Explanation of Solution

Given information:

Let Miami be point M

Bermuda be point B

San Juan be point S.

  $\begin{array}{l}M={53}^0\\ b=1038mi\\ m=965mi\end{array}$

Given

  $\begin{array}{l}M={53}^0\\ b=1038mi\\ m=965mi\end{array}$

We know

Using law of sines

We get

  $\begin{array}{l}\frac{\sin M}{m}=\frac{\sin B}{b}\\ \Rightarrow \frac{\sin {53}^0}{965}=\frac{\sin B}{1038}\\ \Rightarrow \sin B=\frac{1038\sin {53}^0}{965}\\ \Rightarrow \sin B=\frac{1038\left(0.799\right)}{965}\\ \Rightarrow \sin B=\frac{829.362}{965}\\ \Rightarrow \sin B=0.86\\ \Rightarrow B={\sin }^{-1}\left(0.86\right)\\ \Rightarrow B={59}^0\end{array}$

Sum of angles of a triangle is ${180}^0$

  $\begin{array}{l}\Rightarrow B+M+S={180}^0\\ \Rightarrow {59}^0+{53}^0+S={180}^0\\ \Rightarrow {112}^0+S={180}^0\\ \Rightarrow S={180}^0-{112}^0\\ \Rightarrow S={68}^0\end{array}$

Using law of sines

We get

  $\begin{array}{l}\frac{\sin M}{m}=\frac{\sin S}{s}\\ \Rightarrow \frac{\sin {53}^0}{965}=\frac{\sin {68}^0}{s}\\ \Rightarrow s=\frac{965\sin {68}^0}{\sin {53}^0}\\ \Rightarrow s=\frac{965\left(0.9272\right)}{0.7986}\\ \Rightarrow s=\frac{894.748}{0.7986}\\ \Rightarrow s=1120.4\end{array}$

Therefore,

Distance between Miami and Bermuda is $1120.4$ miles

b.

To determine

To find: Area of Bermuda Triangle.

Answer to Problem 40PPS

Area of Bermuda Triangle $=A=464,366.1$ square miles.

Explanation of Solution

Given information:

Let Miami be point M

Bermuda be point B

San Juan be point S.

  $\begin{array}{l}M={53}^0\\ b=1038mi\\ m=965mi\end{array}$

Concept:

Area of Triangle $=A=\frac{1}{2}\left(bm\sin \left(S\right)\right)$

Given

  $\begin{array}{l}M={53}^0\\ b=1038mi\\ m=965mi\end{array}$

We know

Using law of sines

We get

  $\begin{array}{l}\frac{\sin M}{m}=\frac{\sin B}{b}\\ \Rightarrow \frac{\sin {53}^0}{965}=\frac{\sin B}{1038}\\ \Rightarrow \sin B=\frac{1038\sin {53}^0}{965}\\ \Rightarrow \sin B=\frac{1038\left(0.799\right)}{965}\\ \Rightarrow \sin B=\frac{829.362}{965}\\ \Rightarrow \sin B=0.86\\ \Rightarrow B={\sin }^{-1}\left(0.86\right)\\ \Rightarrow B={59}^0\end{array}$

Sum of angles of a triangle is ${180}^0$

  $\begin{array}{l}\Rightarrow B+M+S={180}^0\\ \Rightarrow {59}^0+{53}^0+S={180}^0\\ \Rightarrow {112}^0+S={180}^0\\ \Rightarrow S={180}^0-{112}^0\\ \Rightarrow S={68}^0\end{array}$

  $\begin{array}{l}A=\frac{1}{2}\left(bm\sin \left(S\right)\right)\\ \Rightarrow A=\frac{1}{2}\left(\left(1038\right)\left(965\right)\sin \left({68}^0\right)\right)\\ \Rightarrow A=\frac{1038\times 965\times \left(0.9272\right)}{2}\\ \Rightarrow A=\frac{928732.252}{2}\\ \Rightarrow A=464,366.1\end{array}$

Therefore,

Area of Bermuda Triangle $=A=464,366.1$ square miles.