Problem 43 - Algebra and Trigonometry Real Mathematics, Real People

Question

A ship is 45 miles east and 30 miles south of port. The captain wants to sail directly to port. What bearing should the captain take?

Step-by-Step Solution

Answer

After doing all the calculations, you should get a bearing of around 56.31° (in degrees) or its equivalents depending on your calculator’s settings.

Step 1: Draw a sketch of the situation

The ship's position can be represented by a point in a Cartesian coordinate system where the east-west direction is aligned with the x-axis and the north-south direction is aligned with the y-axis. The southward distance is negative because it refers to a downward direction. The ship is at the point (45, -30), while the port is at the origin (0,0). Draw a line representing the direct route from the ship to port.

Step 2: Calculate the angle theta

The angle \(\theta\) between the line (ship to port) and x-axis (east) can be calculated using the formula for the arctangent (or inverse tangent) which is \(\theta = arctan(\frac{opposite}{adjacent})\). Here, the 'opposite' is the y-coordinate and the 'adjacent' is the x-coordinate. Calculate this as \(\theta = arctan(\frac{-30}{45})\). Remember to convert your angle from radians to degrees at the end, if it is not already.

Step 3: Determine the bearing

Bearings are traditionally measured in degrees from the north line: 0° means north, 90° means east, and so on, counting clockwise. Here, the angle we've calculated is below (south of) the east line, so this needs to be reflected in the bearing. Hence, the bearing would be \(90 - \theta\) degrees, if \( \theta \) is in degrees.

Key Concepts

Understanding Angle Calculation

In trigonometry, calculating an angle involves determining the measure between two lines or two planes. In the scenario of the ship, this means finding the angle from the east line to the direct path to the port. To achieve this, you can use the arctangent, often referred to as the inverse tangent.

Here's how you do it:

• Identify the 'opposite' and 'adjacent' sides of the right triangle formed between the line toward the port and the eastward path. Here, since the ship is east and south of the port, the east direction (x-axis) is adjacent, and the south direction (y-axis) is opposite.
• Apply the formula for arctangent: \( \theta = \arctan\left(\frac{\text{opposite}}{\text{adjacent}}\right) \). For our ship example, it becomes \( \theta = \arctan\left(\frac{-30}{45}\right) \).

This will give you \( \theta \) in radians, which is often how calculators present results. To convert it to degrees (which is usually more user-friendly), multiply it by \( \frac{180}{\pi} \).

Demystifying Bearing Calculation

Bearings are a way to express direction, often used in navigation. When calculating bearings, you measure from the north direction and turn clockwise.

In our ship's scenario, after determining the angle \( \theta \) to the east, you need to adjust this angle to fit within the bearing system. Since bearings start at zero at the north and move clockwise:

• First, realize that a bearing that is due east would be 90°.
• Since the calculated angle \( \theta \) is not east but south of east, you'll subtract \( \theta \) from 90° to find the actual bearing.

This calculation positions the path taking advantage of the cardinal system, starting from north, moving east, and incorporating any southward motion.

Utilizing Inverse Tangent

The inverse tangent function, known as arctan or \( \tan^{-1} \), is crucial for our angle calculation. In practical terms, it answers the question: "What angle has a tangent equal to a given value?" This is why it's so helpful in navigation and trigonometry.

Here's how it works in the context of our exercise:

• Given the coordinates of a ship (45 east, -30 south), find the slope of the line (which is the tangent) representing the direct path back to the port. This is simply: \( \frac{-30}{45} \).
• Use the inverse tangent function to find the angle from this slope: \( \theta = \arctan\left(\frac{-30}{45}\right) \).

This method is a straightforward way to convert between linear coordinates and angular direction, making it invaluable for precise navigational tasks like the one at hand.