Problem 33
Question
A sailboat set a course of \(\mathrm{N} 25^{\circ} \mathrm{E}\) from a small port along a shoreline that runs north and south. Sometime later the boat overturned and the crew sent out a distress call. They estimated that they were 12 miles away from the nearest harbor, which is 28 miles north of the port they had set sail from. If a rescue team leaves from the harbor, find all possible courses the team must follow in order to reach the overturned sailboat.
Step-by-Step Solution
Verified Answer
The rescue team can follow a course of approximately \( N 12^{\circ} E \) or \( S 12^{\circ} W\).
1Step 1: Understand the problem
The problem involves finding the possible direction(s) a rescue team must take from a harbor to an overturned sailboat, which is 12 miles away from the nearest harbor. The boat's initial course was \(N 25^{\circ} E\), and the harbor lies 28 miles directly north of the sailboat's original position.
2Step 2: Analyze the sailboat's route
The sailboat's course \(N 25^{\circ} E\) means it was traveling 25 degrees east of due north. This sets up a right triangle where the sailboat's path is the hypotenuse. Let the port be point P, and the sailboat’s overturned location as S.
3Step 3: Calculate the sailboat's position
Using trigonometric ratios, calculate how far east and north the sailboat traveled from the port:- Northward distance: \(y = 28 + 12\sin(25^{\circ})\)- Eastward distance: \(x = 12\cos(25^{\circ})\).
4Step 4: Locate the harbor and calculate possible rescue courses
The harbor, point H, is 28 miles north of P. Using the coordinates of H and S, calculate the distance from H to S using the Pythagorean theorem, ensuring it is 12 miles as given.
5Step 5: Use trigonometry to find rescue courses
Determine the angle \( \theta \) at point H that the rescue team must take to reach S:\[ \theta = \tan^{-1}\left(\frac{x}{y}\right) \]This gives two possible courses: one following \( \theta \) direction, and another the supplementary angle (i.e., \( 180^{\circ} - \theta \)). Adjust angles to compass headings.
Key Concepts
Compass HeadingsTrigonometric RatiosRight TriangleAngle Calculation
Compass Headings
Compass headings are essential navigational tools in trigonometry-related problems. They help in determining a specific direction by combining the four cardinal directions: North, South, East, and West. In the context of this problem, the sailboat set a course at an angle denoted as \(N 25^\circ E\). This notation means the boat was traveling 25 degrees towards the east, starting from a directly north direction.
To understand compass headings better, you can use:
To understand compass headings better, you can use:
- Cardinal Points: N, S, E, and W which represent primary directions.
- Intercardinal Points: Like NE, SE, NW, and SW, which are in between the cardinal points.
- Degrees: Represent how many degrees an angle deviates from a cardinal point. Here, 25 degrees east of north.
Trigonometric Ratios
Trigonometric ratios are foundations for solving problems involving right triangles. They link angles to side lengths. For right triangles, the three primary trigonometric ratios are sine (sin), cosine (cos), and tangent (tan). In this problem, we utilized these ratios extensively to find the sailboat's northward and eastward displacement.
Here's a snapshot of how these ratios work:
Here's a snapshot of how these ratios work:
- Sine (sin): \( \sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}} \)
- Cosine (cos): \( \cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}} \)
- Tangent (tan): \( \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} \)
Right Triangle
The configuration of the problem forms a right triangle. A right triangle is a triangle where one angle measures exactly 90 degrees. This setup assists in leveraging trigonometric ratios effectively to solve problems.
The sailboat's course began at a port and progressed 25 degrees east of north. This established a right triangle with:
The sailboat's course began at a port and progressed 25 degrees east of north. This established a right triangle with:
- Hypotenuse: The path of the sailboat.
- Opposite side: The distance it moved east.
- Adjacent side: The distance it moved north.
Angle Calculation
Calculating angles in a right triangle involves understanding the relationship between the triangle's sides. With trigonometric functions, you can find any missing angle if you have sufficient side information.
In the exercise, the angle calculation focused on finding the course a rescue team must take from the harbor to the sailboat. The angle \( \theta \) is determined using:\[\theta = \tan^{-1}\left(\frac{\text{Eastward Distance} (x)}{\text{Northward Distance} (y)}\right)\]This inverse tangent function determines \( \theta \) based on the opposite and adjacent sides’ ratios. Plus, the problem suggests considering a supplementary angle \( (180^\circ - \theta) \) to reflect navigation possibilities.
Angle calculations, with their practical applications, help solve puzzles involving navigation, such as finding a directional path as seen with the rescue team's task.
In the exercise, the angle calculation focused on finding the course a rescue team must take from the harbor to the sailboat. The angle \( \theta \) is determined using:\[\theta = \tan^{-1}\left(\frac{\text{Eastward Distance} (x)}{\text{Northward Distance} (y)}\right)\]This inverse tangent function determines \( \theta \) based on the opposite and adjacent sides’ ratios. Plus, the problem suggests considering a supplementary angle \( (180^\circ - \theta) \) to reflect navigation possibilities.
Angle calculations, with their practical applications, help solve puzzles involving navigation, such as finding a directional path as seen with the rescue team's task.
Other exercises in this chapter
Problem 33
Find the magnitude of each of the following vectors. $$\mathbf{W}=-\mathbf{i}-2 \mathbf{j}$$
View solution Problem 33
We know from this section that the area of any triangle \(A B C\) is given by Arca \(=\frac{1}{2} b c \sin A=\frac{1}{2} a c \sin B=\frac{1}{2} a b \sin C\) Use
View solution Problem 34
Angle of Elevation A 155 -foot antenna is on top of a tall building. From a point on the ground, the angle of elevation to the top of the antenna is \(28.5^{\ci
View solution Problem 34
Show that each pair of vectors is perpendicular. \(-4 \mathbf{i}-3 \mathbf{j}\) and \(6 \mathbf{i}-8 \mathbf{j}\)
View solution