A rocket fired straight up is tracked by an observer on the ground a mile away.
a)Show that when the angle of elevation is theta, the height of the rocket in feet is h=5280tan(theta).
To solve this problem, we can use trigonometry. Let's consider the right triangle formed by the observer, the height of the rocket, and the ground.
In the right triangle, the angle of elevation (θ) is the angle between the ground and the line of sight from the observer to the rocket. The opposite side of this angle is the height of the rocket (h), and the adjacent side of this angle is the horizontal distance from the observer to the rocket, which is 1 mile or 5280 feet.
Using the tangent function, we can write:
tan(θ) = h / 5280
To find the value of h, we need to isolate it. Multiplying both sides of the equation by 5280 gives:
5280 * tan(θ) = h
Therefore, the height of the rocket in feet (h) can be expressed as:
h = 5280 * tan(θ)
This equation shows that when the angle of elevation is θ, the height of the rocket in feet is 5280 times the tangent of the angle.
To show that the height of the rocket in feet is given by h = 5280tan(θ), we can make use of the properties of trigonometry.
Let's set up a right triangle. The vertical leg of the triangle represents the height of the rocket (h), and the horizontal leg represents the distance between the observer and the rocket (1 mile or 5280 feet). The hypotenuse of the triangle will be the path of the rocket.
The angle of elevation, θ, is the angle between the horizontal leg and the hypotenuse.
Using the definition of tangent, we know:
tan(θ) = opposite/adjacent.
The opposite side of the triangle is the height of the rocket (h), and the adjacent side is the distance between the observer and the rocket (5280 feet).
Therefore, we have:
tan(θ) = h/5280
Now we can rearrange the equation to solve for h:
h = 5280tan(θ)
We have shown that the height of the rocket, h, in feet is given by h = 5280tan(θ).