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Chapter 3: Problem 7

A radar antenna is tracking a satellite orbiting the earth. At a certain time, the radar screen shows the satellite to be \(162 \mathrm{~km}\) away. The radar antenna is pointing upward at an angle of \(62.3^{\circ}\) from the ground. Find the \(x\) and \(y\) components (in \(\mathrm{km}\) ) of the position of the satellite.

Short Answer

Expert verified
\(x \approx 75.62\, \text{km}\) and \(y \approx 142.83\, \text{km}\).

Step by step solution

01

Understand the Problem

We have a right triangle, where the hypotenuse is the distance of the satellite from the radar, which is 162 km. The angle between the ground and the radar beam is 62.3掳. We aim to find the horizontal (\(x\)) and vertical (\(y\)) components of this position.
02

Calculate the Horizontal Component

The horizontal component (x) can be found using the cosine function. \( x = \text{{hypotenuse}} \times \cos(\text{{angle}}) = 162 \times \cos(62.3掳) \).Calculate \(\cos(62.3掳)\) and multiply by 162 to find \(x\).
03

Calculate the Vertical Component

The vertical component (y) can be found using the sine function. \( y = \text{{hypotenuse}} \times \sin(\text{{angle}}) = 162 \times \sin(62.3掳) \).Calculate \(\sin(62.3掳)\) and multiply by 162 to find \(y\).
04

Solve and Conclude

Calculate the results: \( x = 162 \times \cos(62.3掳) \approx 75.62 \) km and \( y = 162 \times \sin(62.3掳) \approx 142.83 \) km. These are the horizontal and vertical components, respectively.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometry Concepts in Action
Trigonometry is a branch of mathematics that studies the relationships between the angles and sides of a triangle. It's beautiful in its simplicity and power. It helps solve many real-world problems, much like the task of finding the position of a satellite.
In trigonometry, two key functions you often encounter are sine (sin) and cosine (cos). These functions make it easier to break down angles into dimensions that we can measure easily.
  • Sine Function: The sine of an angle in a right triangle is the ratio of the opposite side to the hypotenuse. It's useful for finding vertical heights.
  • Cosine Function: The cosine of an angle in a right triangle is the ratio of the adjacent side to the hypotenuse. It's used to determine horizontal distances.
When you apply these functions to a satellite tracking problem, you convert the angle and hypotenuse into the parts you can measure on the earth's surface. Thus, by just knowing one angle and one distance, trigonometry opens up the whole picture!
Understanding Right Triangles
A right triangle is special because one of its angles is exactly 90 degrees. This creates a world of mathematical possibilities, making calculations easier.
In the context of satellite tracking, visualizing the scenario as a right triangle simplifies the calculation of a satellite鈥檚 exact position. Here's why:
  • The hypotenuse is the known distance, like 162 km in the exercise. This is the line from the radar to the satellite.
  • The angle, such as 62.3掳, is formed by the line from the radar pointing upward and the horizontal ground.
Given these two pieces of information, you can precisely find the other two sides of the triangle. The horizontal and vertical lines (forming the triangle's legs) represent the satellite's position in two-dimensional space.
Satellite Tracking and Component Analysis
Satellite tracking is a fascinating application of real-world data working with mathematical concepts. As satellites orbit Earth, professionals use mathematical models to pinpoint their positions at any given time.
Using trigonometry within the right triangular model, these satellite positions are determined with ease:
  • Horizontal Component: This is the ground distance and can be logged using the cosine function for accuracy.
  • Vertical Component: This is the elevation from the ground, identified using the sine function to access its height.
In a satellite tracking task, breaking down the position into components allows us to clearly see how far the satellite is in each direction. It鈥檚 like having a map with 'x' and 'y' coordinates. Trigonometry helps change complex orbital paths into simple, understandable segments for precise tracking and analysis.

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Most popular questions from this chapter

A speed ramp at an airport is a moving conveyor belt on which you can either stand or walk. It is intended to get you from place to place more quickly. Suppose a speed ramp is \(120 \mathrm{~m}\) long. When you walk at a comfortable speed on the ground, you cover this distance in \(86 \mathrm{~s}\). When you walk on the speed ramp at this same comfortable speed, you cover this distance in 35 s. Determine the speed at which the speed ramp is moving relative to the ground.

A Coast Guard ship is traveling at a constant velocity of \(4.20 \mathrm{~m} / \mathrm{s},\) due east, relative to the water. On his radar screen the navigator detects an object that is moving at a constant velocity. The object is located at a distance of \(2310 \mathrm{~m}\) with respect to the ship, in a direction \(32.0^{\circ}\) south of east. Six minutes later, he notes that the object's position relative to the ship has changed to \(1120 \mathrm{~m}, 57.0^{\circ}\) south of west. What are the magnitude and direction of the velocity of the object relative to the water? Express the direction as an angle with respect to due west.

The ball is \(26.9 \mathrm{~m}\) from the goalpost. The ball is kicked with an initial velocity of \(19.8 \mathrm{~m} / \mathrm{s}\) at an angle \(\theta\) above the ground. Between what two angles, \(\theta_{1}\) and \(\theta_{2}\), will the ball clear the \(2.74\) -m-high crossbar? (Hint: The following trigonometric identities may be useful: \(\sec \theta=1 /(\cos \theta)\) and \(\sec ^{2} \theta=1+\tan ^{2} \theta\)

Two cannons are mounted as shown in the drawing and rigged to fire simultaneously. They are used in a circus act in which two clowns serve as human cannonballs. The clowns are fired toward each other and collide at a height of \(1.00 \mathrm{~m}\) above the muzzles of the cannons. Clown A is launched at a \(75.0^{\circ}\) angle, with a speed of \(9.00 \mathrm{~m} / \mathrm{s}\). The horizontal separation between the clowns as they leave the cannons is \(6.00 \mathrm{~m} .\) Find the launch speed \(v_{0 B}\) and the launch angle \(\theta_{B}\left(>45.0^{\circ}\right)\) for clown B.

The punter on a football team tries to kick a football so that it stays in the air for a long "hang time." If the ball is kicked with an initial velocity of \(25.0 \mathrm{~m} / \mathrm{s}\) at an angle of \(60.0^{\circ}\) above the ground, what is the "hang time"?
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