Problem 39 - Algebra and Trigonometry

Question

Two straight roads diverge at an angle of \(65^{\circ} .\) Two cars leave the intersection at \(2 : 00\) P.M., one traveling at 50 \(\mathrm{mi} / \mathrm{h}\) and the other at 30 \(\mathrm{mi} / \mathrm{h} .\) How far apart are the cars at \(2 : 30 \mathrm{P.M.?}\)

Step-by-Step Solution

Answer

The cars are approximately 23.08 miles apart.

Step 1: Determine the time interval

The cars travel from 2:00 PM to 2:30 PM, which is a half-hour (0.5 hours).

Step 2: Calculate the distance each car travels

For the first car traveling at 50 mi/h, the distance is \(50 \text{ miles/hour} \times 0.5 \text{ hours} = 25 \text{ miles}\). For the second car traveling at 30 mi/h, the distance is \(30 \text{ miles/hour} \times 0.5 \text{ hours} = 15 \text{ miles}\).

Step 3: Identify the triangle formed

The car paths form a triangle with sides of 25 miles and 15 miles, and an included angle of \(65^{\circ}\).

Step 4: Apply the Law of Cosines to find the distance between the cars

Using the Law of Cosines: \( c^2 = a^2 + b^2 - 2ab \cos(C)\) where \(a = 25\), \(b = 15\), and \(C = 65^{\circ}\). Substitute the values: \[c^2 = 25^2 + 15^2 - 2 \times 25 \times 15 \times \cos(65^{\circ}) \]Compute: \(25^2 = 625\), \(15^2 = 225\), and \(\cos(65^{\circ}) \approx 0.422618\), \\(c^2 = 625 + 225 - 2 \times 25 \times 15 \times 0.422618\); calculate \(2 \times 25 \times 15 \times 0.422618 \approx 317.0\); so \(c^2 \approx 625 + 225 - 317.0\).

Step 5: Find the exact distance between the cars

Simplify the expression \(c^2 = 850 - 317\), so \(c^2 = 533\). Take the square root of both sides: \(c \approx \sqrt{533} \approx 23.08\). Thus, the cars are approximately 23.08 miles apart.

Key Concepts

Angle of Divergence

When two paths meet and then start to separate, the angle formed between them is known as the "angle of divergence." This angle is crucial in determining the relationship and distance between the paths over time. In this problem, the roads diverge at an angle of \(65^{\circ}\).
The angle of divergence is a key piece of information in trigonometry when dealing with triangles formed by such diverging paths. It helps in calculating distances and interpreting relationships between those paths.
This is because the angle affects how we apply trigonometric principles like the Law of Cosines. The greater the angle, the further apart the paths or objects at the same distance along the paths. This is why the angle matters so much in distance calculations.

Distance Calculation

Calculating distance between two points often involves geometry and trigonometry, especially when those points move from a common starting location and diverge like the cars in this problem. For these cars:
- One car travels 50 miles/hour, and the other travels 30 miles/hour. - They both travel for 0.5 hours, leading to the first car traveling 25 miles and the second car 15 miles.
To find out how far they are from each other after diverging, we use a special rule called the Law of Cosines. The Law of Cosines helps when you know two sides of a triangle and the included angle, just like our situation. It is given by the formula: \[c^2 = a^2 + b^2 - 2ab \cos(C)\]
Here, the sides \(a\) and \(b\) are 25 miles and 15 miles, and \(C\) is \(65^{\circ}\). By substituting these values into the formula, we can solve for \(c\) to get the distance between the cars.

Triangle Properties

The triangle formed by the paths of the cars is an example of how triangles can represent real-world situations. Some basic properties of triangles help us understand their shape and dimensions:
- **Sides:** In a triangle, the sum of the lengths of any two sides must be greater than the length of the third side. This condition is always satisfied in a well-defined triangle formed by diverging paths.- **Angles:** The sum of angles in any triangle is always \(180^{\circ}\). Here, we focused on the included angle which was given, simplifying our task.
These properties are fundamental to applying the Law of Cosines, which resolved our problem. The formula connected the known side lengths and included angle to find the third side, giving us the exact distance between the cars. The insight here is that properties of triangles underlie common formulas that solve complex problems in simpler steps.