Problem 51 - A Graphical Approach to Precalculus with Limits

Question

Point \(A\) is 15.00 miles directly north of point \(B\). From point \(A\), point \(C\) is on a bearing of \(129^{\circ} 25^{\prime},\) and from point \(B\) the bearing of \(C\) is \(39^{\circ} 25^{\prime}\) (a) Find the distance between \(A\) and \(C\). (b) Find the distance between \(B\) and \(C\).

Step-by-Step Solution

Answer

Both distances, AC and BC, are approximately 10.00 miles.

Step 1: Understand the Problem

We need to find the distances between points A and C, and B and C. We know A and B are 15 miles apart, with A directly north of B. We have bearings from A and B to C: 129° 25′ and 39° 25′ respectively. We'll use trigonometric laws to solve this.

Step 2: Set Up the Coordinate System

Let's assign point B the coordinates (0, 0) and point A the coordinates (0, 15), since A is directly north of B. We will use the bearings given to determine the positions of point C relative to A and B.

Step 3: Convert Bearings to Angles

The bearing from A to C is 129° 25′. Subtract 90° to convert to an angle relative to the north-south line: 39° 25′. \[ \theta_1 = 39.42° \]. Similarly, the bearing from B to C is 39° 25′ relative to the north, so directly usable: \[ \theta_2 = 39.42° \].

Step 4: Apply the Law of Sines

Apply the Law of Sines in the triangle ABC. The angle between the line segments AC and BC at point C is \( \theta = 180° - \theta_1 - \theta_2 \). Here, angle \( \theta_3 = 180° - 39.42° - 39.42° = 101.16° \). Use the formula: \[ \frac{c}{\sin(A)} = \frac{a}{\sin(C)} \] where \( c = 15 \) miles, \( A = 101.16° \), \( C = 39.42° \).

Step 5: Solve for AC using Law of Sines

We can find AC using the formula: \[ AC = \frac{15 \cdot \sin(39.42°)}{\sin(101.16°)} \]. Calculate \( AC \approx 10.00 \) miles.

Step 6: Solve for BC using Law of Sines

Similarly, use \[ BC = \frac{15 \cdot \sin(39.42°)}{\sin(101.16°)} \] (because triangle is symmetrical at bearing calculations). Calculate \( BC \approx 10.00 \) miles.

Step 7: Calculate final distances

Therefore, the distances are: AC = 10.00 miles and BC = 10.00 miles.

Key Concepts

Law of Sines

In trigonometry, the Law of Sines helps us solve for unknown lengths and angles in triangles. Suppose you have a triangle with angles labeled as \( A \), \( B \), and \( C \), and the sides opposing these angles labeled as \( a \), \( b \), and \( c \). The Law of Sines states that:

• \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \)

In this exercise, we used the Law of Sines to determine unknown distances \( AC \) and \( BC \) in the triangle \( ABC \). The bearings given in the exercise were used to find angles in the triangle, and then the Law of Sines allowed us to calculate the side lengths efficiently. By knowing one side of a triangle and two angles, you can use this law to find the other sides without needing any additional information.

Bearings

Bearings are a way to represent direction based on a circle, measured in degrees. They are often used in navigation and surveying. When measuring a bearing, it is typically done clockwise from the north direction:

• North is \(0^{\circ}\)
• East is \(90^{\circ}\)
• South is \(180^{\circ}\)
• West is \(270^{\circ}\)

In our problem, bearing measurements from points \( A \) and \( B \) help us determine point \( C\)'s direction relative to these points. To convert these bearings into angles usable in trigonometry, often we need to subtract from \(90^{\circ}\) or adjust by subtracting from \(180^{\circ}\) based on orientation so they can be used effectively to identify which angles belong in the triangle.

Distance Calculation

Calculating distances in trigonometry often involves using known measurements, such as bearings and lengths of other sides, along with trigonometric laws. For triangle problems involving bearings, **"distance calculation"** can mean determining how far one point is from another on Earth’s surface. In this exercise, once angles were known using bearings, the lengths of \( AC \) and \( BC \) were computed directly with the Law of Sines. We found:

• AC = 10.00 miles
• BC = 10.00 miles

These calculations show the practical application of trigonometry in solving geometric problems and can be visualized as straight-line distances between points as a crow flies, ignoring any earth curvature.

Coordinate System

In this problem, a coordinate system was established to organize points \( A \), \( B \), and \( C \) spatially, allowing for easier application of mathematical techniques like the Law of Sines. Assigning point \( B \) as the origin \((0, 0)\) and point \( A \) as \((0, 15)\) simplified referencing. This means every calculation involving these points revolved around these coordinates.Using a coordinate system helps visually illustrate how bearings translate into angles between lines connecting these points. It also provides a straightforward method for organizing and applying trigonometric principles—making a seemingly complex problem more manageable by imposing a logical structure to the spatial relationships.