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Lapatulllka [165]
2 years ago
10

A ship leaves port at noon and has a bearing of S29oW. The ship sails at 20 knots. How many nautical miles south and how many na

utical miles west will the ship have traveled by 6:00 PM
Mathematics
1 answer:
ira [324]2 years ago
7 0

Answer:

Approximately 58.2\; \text{nautical miles} (assuming that the bearing is {\rm S$29^{\circ}$W}.)

Step-by-step explanation:

Let v denote the speed of the ship, and let t denote the duration of the trip. The magnitude of the displacement of this ship would be v\, t.

Refer to the diagram attached. The direction {\rm S$29^{\circ}$W} means 29^{\circ} west of south. Thus, start with the south direction and turn towards west (clockwise) by 29^{\circ} to find the direction of the displacement of the ship.

The hypothenuse of the right triangle in this diagram represents the displacement of the ship, with a length of v\, t. The dashed horizontal line segment represents the distance that the ship has travelled to the west (which this question is asking for.) The angle opposite to that line segment is exactly 29^{\circ}.

Since the hypotenuse is of length v\, t, the dashed line segment opposite to the \theta = 29^{\circ} vertex would have a length of:

\begin{aligned}& \text{opposite (to $\theta$)} \\ =\; & \text{hypotenuse} \times \frac{\text{opposite (to $\theta$)}}{\text{hypotenuse}} \\ =\; & \text{hypotenuse} \times \sin (\theta) \\ =\; & v\, t \, \sin(\theta) \\ =\; & v\, t\, \sin(29^{\circ})\end{aligned}.

Substitute in \begin{aligned} v &= 20\; \frac{\text{nautical mile}}{\text{hour}}\end{aligned} and t = 6\; \text{hour}:

\begin{aligned} & v\, t\, \sin(29^{\circ}) \\ =\; & 20\; \frac{\text{nautical mile}}{\text{hour}} \times 6\; \text{hour} \times \sin(29^{\circ}) \\ \approx\; & 58.2\; \text{nautical mile}\end{aligned}.

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