Problem 29 - College Trigonometry

Question

Carl spies a potential Sasquatch nest at a bearing of \(\mathrm{N} 10^{\circ} \mathrm{E}\) and radios Jeff, who is at a bearing of \(\mathrm{N} 50^{\circ} \mathrm{E}\) from Carl's position. From Jeff's position, the nest is at a bearing of \(\mathrm{S} 70^{\circ} \mathrm{W}\). If Jeff and Carl are 500 feet apart, how far is Jeff from the Sasquatch nest? Round your answer to the nearest foot.

Step-by-Step Solution

Answer

Jeff is approximately 643 feet from the Sasquatch nest.

Step 1: Draw the Problem

Visualize the problem by drawing a triangle where Carl's position, Jeff's position, and the nest are the vertices. Label Carl's position as C, Jeff's position as J, and the nest as N.

Step 2: Identify the Bearings

Convert the given bearings into angles relative to the lines of the triangle. Carl sees the nest at \(\mathrm{N}10^{\circ}\mathrm{E}\), and Jeff's position from Carl's is \(\mathrm{N}50^{\circ}\mathrm{E}\), forming an internal angle \(\angle CJN=40^{\circ}\). The nest from Jeff is at \(\mathrm{S}70^{\circ}\mathrm{W}\), which translates to \(110^{\circ}\) as an internal angle at J.

Step 3: Calculate the Third Angle of the Triangle

Using the fact that the sum of angles in a triangle is \(180^{\circ}\), calculate \(\angle CNJ\) as follows:\[\angle CNJ = 180^{\circ} - 40^{\circ} - 110^{\circ} = 30^{\circ}\].

Step 4: Apply the Law of Sines

Use the known angles and the distance between Carl and Jeff (500 feet) to apply the Law of Sines:\[\frac{CN}{\sin(40^{\circ})} = \frac{500}{\sin(30^{\circ})}\].Solve for \(CN\):\[CN = \frac{500 \times \sin(40^{\circ})}{\sin(30^{\circ})}\].

Step 5: Solve for Distance CN

Calculate \(CN\) using the values for the sine of the angles:\[CN = \frac{500 \times 0.6428}{0.5} = 642.8\].Thus, the distance from Jeff to the nest is approximately 643 feet.

Key Concepts

Bearings and navigation

When it comes to navigation, bearings are crucial. They are expressed in degrees and determine the direction from one point to another. Bearings are typically measured clockwise starting from the North direction. This straightforward method allows navigators to pinpoint a location accurately.

In the problem involving Carl and Jeff, we deal with bearings to determine positions relative to each other. Carl sees the nest bearing at \(\mathrm{N}10^{\circ}\mathrm{E}\), meaning the nest is 10 degrees towards the East from due North. Jeff's position from Carl is at \(\mathrm{N}50^{\circ}\mathrm{E}\), and from Jeff’s perspective, the nest is at \(\mathrm{S}70^{\circ}\mathrm{W}\). Understanding these bearings helps define the angles at the vertices of the triangle formed by Carl, Jeff, and the nest.

The key takeaway on bearings is that they help set the stage in navigation problems by accurately defining directional relationships. This becomes essential when trying to compute distances or angles in trigonometric problems.

Law of Sines

The Law of Sines is foundational in solving triangles, particularly when we know some angles and one side, but need to find the unknown sides or angles.
It states that:

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

where \(a\), \(b\), and \(c\) are the lengths of the sides opposite angles \(A\), \(B\), and \(C\), respectively. This relationship becomes practical when working with oblique (non-right) triangles.

In the case of Carl and Jeff's problem, by applying the Law of Sines, we can find the unknown distance from Jeff to the nest. With the angles \(\angle CJN = 40^{\circ}\), \(\angle CNJ = 30^{\circ}\), and \(\angle JCN = 110^{\circ}\), and knowing one side (Carl to Jeff distance of 500 feet), we can solve for the unknown side using:

\[CN = \frac{500 \times \sin(40^{\circ})}{\sin(30^{\circ})}\]

Understanding this law is crucial for solving a variety of trigonometry problems involving non-right triangles.

Angle calculation in triangles

Triangles are a common shape in geometry, and understanding how to calculate angles within them is vital. A triangle's internal angles always sum to \(180^{\circ}\).

For solving problems like the one Carl and Jeff faced, knowing two angles allows you to easily find the third. Subtract the sum of the known angles from \(180^{\circ}\).

Consider their problem:

• The angle from Carl to Jeff to the nest \(\angle CJN\) is \(40^{\circ}\)
• The angle from Jeff relative to his position to the nest \(\angle JCN\) is \(110^{\circ}\)

To find \(\angle CNJ\), you use:

\[\angle CNJ = 180^{\circ} - 40^{\circ} - 110^{\circ} = 30^{\circ}\]

This straightforward method of calculating angles is applicable in many geometric problems, greatly assisting in the visualization and solving of complex trigonometric exercises.