Problem 41 - Algebra and Trigonometry

Question

A pilot flies in a straight path for 1 h 30 min. She then makes a course correction, heading \(10^{\circ}\) to the right of her original course, and flies 2 \(\mathrm{h}\) in the new direction. If she maintains a constant speed of \(625 \mathrm{mi} / \mathrm{h},\) how far is she from her starting position?

Step-by-Step Solution

Answer

The pilot is approximately 367.5 miles from her starting position.

Step 1: Calculate Distances Traveled

First, determine the distances the pilot traveled based on the given times and speed. \The speed of the plane is 625 miles/hour. Start by converting 1 hour 30 minutes to hours: \1 hour 30 minutes = 1.5 hours. \Now calculate the distance for each segment: \\( \text{Distance}_1 = 625 \text{ miles/hour} \times 1.5 \text{ hours} = 937.5 \text{ miles} \). \For the second segment: \\( \text{Distance}_2 = 625 \text{ miles/hour} \times 2 \text{ hours} = 1250 \text{ miles} \).

Step 2: Draw the Path and Apply Cosine Law

Imagine the path as creating a triangle, where the pilot changes course after her first flight segment. The original path is one side of the triangle, the new direction creates the second side, and the distance between starting and ending points forms the third side. The angle between the original path and the new path is \( 10^{\circ} \) after the correction. We'll use the cosine law to find the third side of the triangle:\[ c^2 = a^2 + b^2 - 2ab \cos(C) \]where \( a = 937.5 \text{ miles} \), \( b = 1250 \text{ miles} \), and \( C = 10^{\circ} \).

Step 3: Apply the Cosine Law

Substitute the known values into the cosine law formula to calculate the distance from the starting position:\ \[ c^2 = (937.5)^2 + (1250)^2 - 2 \times 937.5 \times 1250 \times \cos(10^{\circ}) \] \Calculate \( (937.5)^2 = 878906.25 \), \( (1250)^2 = 1562500 \), \and \( 2 \times 937.5 \times 1250 = 2343750 \). \Substitute and compute \( c^2 = 878906.25 + 1562500 - 2343750 \times \cos(10^{\circ}) \).

Step 4: Solve for the Distance

Now solve for \( c \): Calculate \( \cos(10^{\circ}) \approx 0.9848 \), \thus the expression becomes:\\[ c^2 = 878906.25 + 1562500 - 2343750 \times 0.9848 \]On simplifying further, you get:\\[ c^2 = 878906.25 + 1562500 - 2307393.125 = 135013.125 \] Now solve for \( c \) by taking the square root: \\[ c = \sqrt{135013.125} \approx 367.5 \text{ miles} \].

Key Concepts

Distance Calculation

In problems involving linear motion, calculating distance is often the first step. Knowing how fast and how long something travels gives you the distance.
To find the distance traveled, use the formula:

• Distance = Speed × Time

For instance, in our exercise, the pilot travels at a constant speed of 625 miles per hour. It's important to keep time units consistent, so we convert minutes to hours first. Here, 1 hour and 30 minutes is converted to 1.5 hours.
For each flight leg:

• The first segment distance is calculated as 625 miles/hour × 1.5 hours = 937.5 miles.
• The second segment, with 2 hours, results in a distance of 1250 miles.

This lays the groundwork for further calculations, as it's crucial in setting up the entire problem.

Triangle Geometry

Once the pilot changes her path, we can visualize her journey as a triangle. Understanding the relationships in triangle geometry helps in calculating the resultant distance. The initial leg and the course-corrected leg form two sides of the triangle, while the straight-line distance from start to finish creates the third side.
To solve these types of scenarios, it's useful to sketch the paths to see how they form a triangle:

• Each leg of the flight comprises two sides of the triangle: the first at 937.5 miles, and the second at 1250 miles.
• The angle between these sides, due to the course correction, is given as 10 degrees.

Understanding this setup makes it easier to apply the cosine law, which connects the sides of a triangle with the cosines of its angles, to find unknown lengths.

Angle Measurement

Angles play a vital role in determining distances in triangle problems. In this scenario, the pilot's path change results in a 10-degree angle between the first and second legs of the flight. This angle is crucial for applying the cosine law, enabling us to compute the direct distance back to the start point.
To use angles effectively:

• Recognize that a small angle like 10 degrees suggests only slight divergence from the original path.
• Ensure that you apply the angle in the cosine law formula accurately:
• \[c^2 = a^2 + b^2 - 2ab \cos(C)\]
• Here, \( C = 10^{\circ} \), and understanding how to convert this angle to cosine is critical, simply use \( \cos(10^{\circ}) \approx 0.9848 \).

This precise measurement allows determining exact distances scientifically and is a fundamental application in trigonometry.