Problem 50 - Algebra and Trigonometry Real Mathematics, Real People

Question

Two ships leave a port at 9 A.M. One travels at a bearing of \(\mathrm{N} 53^{\circ} \mathrm{W}\) at 12 miles per hour, and the other travels at a bearing of \(\mathrm{S} 67^{\circ} \mathrm{W}\) at 16 miles per hour. Approximate how far apart the ships are at noon.

Step-by-Step Solution

Answer

The two ships are around 42 miles apart at noon.

Step 1: Interpret the Bearings and Speeds

The first ship travels at a bearing of N53°W which means it travels 53 degrees west of due north. It is traveling at 12 mph. The other ship travels at a bearing of S67°W which means it travels 67 degrees west of due south. This ship travels at 16 mph.

Step 2: Calculate the Distance Traveled

Since both ships leave the port at 9 A.M. and the time is now noon, three hours have passed. So, the first ship has traveled \(3* 12 = 36\) miles, and the second ship has traveled \(3 * 16 = 48\) miles.

Step 3: Determine the Angle Between the Paths

The angle between north and south is 180 degrees. The first ship has deviated 53 degrees from north towards west, and the second ship has deviated 67 degrees from south towards west. This makes the angle between them as \(180 - 53 - 67 = 60\) degrees.

Step 4: Use the Law of Cosines to Find the Distance

Finally, we can use the law of cosines to calculate the distance between the two ships. The Law of Cosines is \(c = \sqrt{a^2 + b^2 - 2*a*b*Cos(C)}\) where \(C\) is the angle between \(a\) and \(b\). By substituting \(a = 36\), \(b = 48\), and \(C = 60\) degrees into the formula, the distance \(c = \sqrt{36^2 + 48^2 - 2 * 36 * 48 * Cos(60)}\) miles.

Key Concepts

Bearings and Angles

Understanding bearings and angles is crucial when it comes to navigation and distance calculations. In navigation, a bearing is a direction or path along which something moves or along which it lies. Bearings are measured in degrees, with 0° representing north, 90° representing east, 180° representing south, and 270° representing west.
In the context of the problem, the bearing determines the direction each ship travels. The first ship moves at a bearing of N53°W, which means it is 53 degrees left from north, towards the west direction. The second ship moves at a bearing of S67°W, indicating a direction that is 67 degrees west from the due south line.
This concept of bearings, combined with the understanding that the angle between north and south is 180 degrees, helps in later calculating the angle between the two paths of the ships. This is key for applying the Law of Cosines. Think of it like a compass guiding the ships on their courses, with each bearing marking out their distinct path across the sea.

Distance Calculation

Calculating the distance the ships travel involves understanding speed and time. The formula for distance is:

• Distance = Speed × Time

In this exercise, both ships start their journey at 9 A.M. and we want to know their position at noon. This gives us a total travel time of 3 hours.
For the first ship:

• Speed = 12 miles per hour
• Time = 3 hours
• Distance traveled = 12 mph × 3 hours = 36 miles

For the second ship:

• Speed = 16 miles per hour
• Time = 3 hours
• Distance traveled = 16 mph × 3 hours = 48 miles

These calculations give us the lengths of two sides of a triangle formed by the paths of the ships and the point of departure.

Trigonometry Applications

Trigonometry, especially the Law of Cosines, allows us to find unknown sides in triangles when certain angles and sides are known.

• The Law of Cosines formula is: \[c = \sqrt{a^2 + b^2 - 2ab \cdot \cos(C)}\]

Here:

• \(a = 36\) miles (distance traveled by the first ship)
• \(b = 48\) miles (distance traveled by the second ship)
• \(C = 60°\) (angle between the paths of the two ships)

Substituting these values into the formula, we calculate: \[c = \sqrt{36^2 + 48^2 - 2 \cdot 36 \cdot 48 \cdot \cos(60°)}\]Using a calculator, you find:

• \(c = \sqrt{1296 + 2304 - 1728}\)
• \(c = \sqrt{1872}\)
• The distance \(c \approx 43.27\) miles

This result is the approximate distance between the two ships at noon, demonstrating how trigonometry helps solve real-world distance problems.