Problem 41
Question
Marine Transportation A ship leaves port at noon and has a bearing of \(S 29^{\circ} \mathrm{W}\). The ship sails at 20 knots. How many nautical miles south and how many nautical miles west does the ship travel by 6: 00 p.M.?
Step-by-Step Solution
Verified Answer
The ship travels approximately 57.82 nautical miles south and 103.92 nautical miles west.
1Step 1: Determine the total distance traveled
Since speed is distance over time, and we know both the speed (20 knots) and the time (6 hours), we can multiply these together to get the total distance the ship has traveled. Therefore, the total distance covered by the ship is \(20 \times 6 = 120\) nautical miles.
2Step 2: Calculate the distance traveled in the south direction
We can use the sine of the angle to determine this distance. The sine of an angle in a right triangle is the length of the opposite side divided by the length of the hypotenuse. So, since the bearing is \(29^{\circ}\) west of South, we can say that the opposite side is the distance in the south direction and the hypotenuse is the total distance traveled. That gives us \(\sin(29^{\circ}) = \) Distance South / 120. Solving for Distance South gives us \(\sin(29^{\circ}) \times 120 \approx 57.82 \) nautical miles.
3Step 3: Calculate the distance traveled in the west direction
This is similar to step 2, but instead of using the sine, we use the cosine of the angle because in our case the west distance is adjacent to the angle. This leads to the equation: \(\cos(29^{\circ}) = \) Distance West / 120. Solving this for Distance West gives us \(\cos(29^{\circ}) \times 120 \approx 103.92 \) nautical miles.
Key Concepts
BearingNautical MilesRight TriangleSine and Cosine
Bearing
In navigation, bearing is a crucial concept that helps us understand direction. Bearing is an angle measured clockwise from the north direction to the direction we are interested in. For example, if a ship has a bearing of \( S 29^{\circ} \mathrm{W} \), it means the ship is heading 29 degrees towards the west, starting from the south.
This can also be understood as moving southward, slightly veering westward. Bearings are important as they assist navigators in keeping track of travel directions, especially at sea, where landmarks are scarce.
This can also be understood as moving southward, slightly veering westward. Bearings are important as they assist navigators in keeping track of travel directions, especially at sea, where landmarks are scarce.
Nautical Miles
Nautical miles are a unit of measurement used in maritime and air navigation to measure distance. A nautical mile is based on the circumference of the Earth and is equal to one minute of latitude, approximately 1.15078 miles.
The importance of using nautical miles lies in the fact that they are directly related to degrees of latitude and longitude, making them extremely useful for calculations based on the Earth's geography. For example, knowing a ship traveled 120 nautical miles, as in the exercise, allows navigators to determine precisely how far and in which direction the ship has traveled in relation to the Earth's grid.
The importance of using nautical miles lies in the fact that they are directly related to degrees of latitude and longitude, making them extremely useful for calculations based on the Earth's geography. For example, knowing a ship traveled 120 nautical miles, as in the exercise, allows navigators to determine precisely how far and in which direction the ship has traveled in relation to the Earth's grid.
Right Triangle
A right triangle is a triangle that has one angle exactly equal to 90 degrees. This type of triangle is fundamental to trigonometry as it helps in solving real-world navigation and engineering problems through relationships between its sides and angles.
In our exercise, the bearing angle creates a scenario where the ship's path can be visualized as a right triangle, with the traveled path as the hypotenuse. The measured angles and sides allow us to apply trigonometric functions like sine and cosine to calculate distances in specific directions.
In our exercise, the bearing angle creates a scenario where the ship's path can be visualized as a right triangle, with the traveled path as the hypotenuse. The measured angles and sides allow us to apply trigonometric functions like sine and cosine to calculate distances in specific directions.
Sine and Cosine
Sine and cosine are trigonometric functions that relate the angles and sides of a right triangle. In a right triangle, sine of an angle is the ratio of the length of the opposite side to the hypotenuse, while cosine is the ratio of the length of the adjacent side to the hypotenuse.
In the problem, using "sine" and "cosine" allowed us to break down the ship's journey accurately. We used the sine function to calculate the distance traveled in the south direction, as this distance is opposite to the given angle. Similarly, the cosine function was used to determine the westward distance, as this distance is adjacent to the angle. These functions simplify complex navigational calculations by breaking the vectors of travel into understandable components.
In the problem, using "sine" and "cosine" allowed us to break down the ship's journey accurately. We used the sine function to calculate the distance traveled in the south direction, as this distance is opposite to the given angle. Similarly, the cosine function was used to determine the westward distance, as this distance is adjacent to the angle. These functions simplify complex navigational calculations by breaking the vectors of travel into understandable components.
Other exercises in this chapter
Problem 40
Use a graphing utility to graph \(f\) and \(g\) in the same viewing window. (Include two full periods.) Make a conjecture about the functions. $$\begin{aligned}
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Use the angle-conversion capabilities of a graphing utility to convert the angle measure to \(\mathbf{D}^{\circ} \mathbf{M}^{\prime} \mathbf{S}^{\prime \prime}\
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Complete the identity. $$\sec \theta=\frac{1}{\square}$$
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Compare the graph of the function with the graph of \(f(x)=\arctan x\) \(g(x)=\arctan x+1\)
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