Chapter 6.1, Problem 10QR

Solution

To calculate: The distance covered after traveling $8$ hours from Port Norfolk and also calculate the bearing of boat.

Distance covered: $254.1\text{ nautical miles}$

Bearing: ${95.4}^{\circ }$

Given:

Average of ship $=42\text{ knots}$

Traveling duration-1: $=3\text{ hours}$

Bearing1 $={40}^{\circ }$

Traveling duration-2 $=5\text{ hr}$

Bearning2 $={125}^{\circ }$

Explanation:

The distance covered in $3$ hours $=42\left(3\right)=126\text{ knots}$

The distance covered in $5$ hours $=42\left(5\right)=210\text{ knots}$

Suppose after $8$ hours the covered distance is $x$ and the bearing is $\theta$ .

Consider the following figure:

  

Use the law of cosine to obtain the value of $x$ .

  $\begin{array}{l}{x}^2={126}^2+{210}^2-2\left(126\right)\left(210\right)\cos \left({40}^{\circ }+{55}^{\circ }\right)\\ x=\sqrt{{126}^2+{210}^2-2\left(126\right)\left(210\right)\cos {95}^{\circ }}\\ x\approx 254.1\text{ nautical miles}\end{array}$

Again, obtain the value of $A$ .

  $\begin{array}{c}\frac{\sin A}{210}=\frac{\sin {95}^{\circ }}{254.1}\\ \sin A=\frac{210\sin {95}^{\circ }}{254.1}\\ A={\sin }^{-1}\left(\frac{210\sin {95}^{\circ }}{254.1}\right)\\ A\approx {55.4}^{\circ }\end{array}$

So, the bearing of the boat is,

  $\begin{array}{l}\theta ={40}^{\circ }+A\\ \theta \approx {40}^{\circ }+{55.4}^{\circ }\\ \theta \approx {95.4}^{\circ }\end{array}$

Conclusion:

So, the distance covered is $254.1\text{ nautical miles}$ and the bearing is ${95.4}^{\circ }$ .