Chapter 4.8, Problem 35E

a.

To determine

How many nautical miles south the ship have travelled by 6:00 p.m.?

Answer to Problem 35E

  $\boxed{104.95}$ nautical miles

Explanation of Solution

Given information:

A ship leaves port at noon and has a bearing of $\text{S 29}°\text{ W}$ . The ship sails at $\text{2}0$ knots.

How many nautical miles south and how many nautical miles west will the ship have travelled by 6:00 p.m.?

Calculation:

The following figure indicates how the ship is moving. The point $P$ shows the ship. We will use the fact that it leaves port at noon, has a bearing $\text{S 29}°\text{ W}$ and that it sails at $\text{2}0$ knots.

Let the vertical distance travelled by the ship, southwards, at 6 P.M. that is after $6$ hours be $s$ .

The bearing of ship is south $29°$ west, which means ship’s sailing at an angle of $209°$ from south direction.

  $270°-209°$

  $=61°$

Also, the distance travelled by the ship sailing at $\text{2}0$ knots after $6$ hours

  $dis\tan ce=speed\times time$

  $=20\times 6$

  $=120$ nautical miles

From the figure, in right $\Delta OWV$ ,

  $\sin 61°=\frac{\text{Opposite}}{\text{Hypotenuse}}$

  $\sin 61°=\frac{s}{120}$

  $120\times \sin 61°=s$

  $120\times 0.8746=s$

  $s=104.95$

Hence, the ship is sailing $\boxed{104.95}$ nautical miles south after $6$ hours.

b.

To determine

How many nautical miles west the ship have travelled by 6:00 p.m.?

Answer to Problem 35E

  $\boxed{58.176}$

Explanation of Solution

Given information:

A ship leaves port at noon and has a bearing of $\text{S 29}°\text{ W}$ . The ship sails at $\text{2}0$ knots.

A t 6:00 p.m., the ship changes course to due west. Find the ship’s bearing and distance from port at 7:00 p.m.

Calculation:

Now, let the horizontal distance travelled by ship, westwards, after $6$ hours be $w$ .

Again, in right $\Delta OWV$ ,

  $\sin 61°=\frac{\text{Opposite}}{\text{Hypotenuse}}$

  $\cos 61°=\frac{w}{120}$

  $120\times \cos 61°=w$

  $120\times 0.4848=w$

  $w=58.176$

Hence, the ship is sailing $\boxed{58.176}$ nautical miles west after $6$ hours journey.