Chapter 5.7, Problem 36E

(a)

To determine

In what direction should the captain turn the ship to arrive at the port?

Answer to Problem 36E

  ${17.2}^{\circ }\text{East of North}$

Explanation of Solution

Given information: Navigation: The captain of the Coast Guard Cutter Pendant plans to sail to a port that is 450 miles away and ${12}^{\circ }$ east of north. The captain first sails the ship due north to check a buoy. He then turns the ship and sails 316 miles to the port.

Calculation:

Observe the given triangle.

The turn angle is opposite to side 450 and 316 is opposite to ${12}^{\circ }$

  $\begin{array}{l}{12}^{\circ }<{90}^{\circ }\\ 316>450\cdot \sin {12}^{\circ }\end{array}$

Case I, two solutions.

  $\begin{array}{l}\frac{316}{\sin {12}^{\circ }}=\frac{450}{\sin \theta }\\ \sin \theta =\frac{450\times \sin {12}^{\circ }}{316}\\ \theta ={\sin }^{-1}\left(\frac{450\times \sin {12}^{\circ }}{316}\right)\approx 17.2221\\ \theta \text{can also be }{\theta }_2={180}^{\circ }-{17.2}^{\circ }={162.8}^{\circ }\end{array}$

The angle of turn is (in terms of triangle angles) obtuse, so, it is

  ${162.8}^{\circ }$ to the current course or in terms of sides of the Earth, this can be expressed as

  ${17.2}^{\circ }\text{East of North}$

(b)

To determine

How many hours will it take to arrive at the turning point if the captain chooses a speed of 23 miles per hour?

Answer to Problem 36E

  $6\text{Hours}$

Explanation of Solution

Given information: Navigation: The captain of the Coast Guard Cutter Pendant plans to sail to a port that is 450 miles away and ${12}^{\circ }$ east of north. The captain first sails the ship due north to check a buoy. He then turns the ship and sails 316 miles to the port.

Calculation:

Observing the given triangle and having calculated the second (turning point) in part (a),

We now find the third angle.

The side opposite is the first leg of the journey,

  $\theta ={180}^{\circ }-\left({162.8}^{\circ }+{12}^{\circ }\right)={5.2}^{\circ }$

For the distance d :

  $\begin{array}{l}\frac{d}{\sin {5.2}^{\circ }}=\frac{316}{\sin {12}^{\circ }}\\ d=\frac{316\times \sin {5.2}^{\circ }}{\sin {12}^{\circ }}\approx 138.30947\end{array}$

The average speed was 23 miles per hour

  $v=\frac{d}{t}\Rightarrow t=\frac{d}{v}=\frac{138.3}{23}\approx 6.013455$

Thus,

6 hours

(c)

To determine

Will the ship reach the port by following this plan, if instead of the plan above, the captain decides to sail 200 miles north, turn through an angle of ${20}^{\circ }$ east of north and then sail along a straight course?

Answer to Problem 36E

No, the ship will not reach the port

Explanation of Solution

Given information: Navigation: The captain of the Coast Guard Cutter Pendant plans to sail to a port that is 450 miles away and ${12}^{\circ }$ east of north. The captain first sails the ship due north to check a buoy. He then turns the ship and sails 316 miles to the port.

Calculation:

The angle of ${12}^{\circ }$ and the distance of 450 miles are still the same as before.

Draw a new triangle with new information and calculate the new two other angles of the triangle.

  $\begin{array}{l}{180}^{\circ }-{20}^{\circ }={160}^{\circ }\\ {180}^{\circ }-\left({160}^{\circ }+{12}^{\circ }\right)=8^{\circ }\end{array}$

Side of 200 miles is opposite to $8^{\circ }$ ,

And x is the 450 miles distance to the port. Or is it?

  $\frac{200}{\sin 8^{\circ }}=\frac{x}{\sin {160}^{\circ }}$

  $\begin{array}{l}x=\frac{200\times \sin {160}^{\circ }}{\sin 8^{\circ }}\approx 491.5032\\ 450\ne 491.5\end{array}$

Thus,

No, the ship will not reach the port