Chapter 6.6, Problem 45E

(a)

To determine

To find: The distance between the fisherman’s home port and Forrest Island.

Answer to Problem 45E

The distance between the fisherman’s home port and Forrest Island is $x\approx 62.6\text{mi}$.

Explanation of Solution

Given:

The below figure shows the distance between the fisherman’s home port and Forrest Island.

Figure (1)

Formula used: Law of Cosines.

$b^2=a^2+c^2-2ac\cos B$.

Calculation:

Draw a line between forrest island and home port.

Figure (2)

Substitute $50$ for $a$, $x$ for $b$, $30$ for $c$ and $100°$ for $B$ in law of cosines,

$\begin{array}{c}{x}^2={\left(50\right)}^2+{\left(30\right)}^2-2\left(50\right)\left(30\right)\cos 100°\\ =2500+900-3000\left(-0.173648\right)\\ =3400+520.944531\\ =3920.944531\end{array}$

Simplify the above expression by taking square root to each side,

$\begin{array}{c}\sqrt{x^2}=\sqrt{3920.944531}\\ x=62.61744531\\ x\approx 62.6\text{mi}\end{array}$

Thus, the distance between the fisherman’s home port and Forrest Island is $x\approx 62.6\text{mi}$.

(b)

To determine

To find: The bearing from Forrest Island back to his home port.

Answer to Problem 45E

The bearing from Forrest Island back to his home port is $\text{S}18.2°\text{E}$.

Explanation of Solution

Given:

The figure,

Figure (1)

Formula used: Law of Sines.

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

Calculation:

Substitute $62.6$ for b, $30$ for $c$ and $100$ for $B$ in Law of Sines ,

$\begin{array}{c}\frac{\sin \left(100\right)}{62.6}=\frac{\sin C}{30}\\ \sin C=\frac{30\left(0.9848\right)}{62.6}\\ \sin C=\frac{\left(29.54423\right)}{62.6}\\ \sin C=0.47195\end{array}$

Simlify the above equation,

$\begin{array}{c}C={\sin }^{-1}\left(0.47195\right)\\ =28.16111\\ C\approx 28.2°\end{array}$

The bearing from Forrest Island to Egg Island is $\text{S}10°\text{W}$.

The bearing from Forrest Island back to his home port is the difference of the angle $C$ and the bearing from Forrest Island to Egg Island,

$\begin{array}{c}\theta =C-10°\\ =28.2°-10°\\ =18.2°\end{array}$

Thus, the bearing from Forrest Island back to his home port is $\text{S}18.2°\text{E}$.