A rocket fired straight up is tracked by an observer on the ground 1 mi away.

(a) Show that when the angle of elevation is 𝜃, the height of the rocket (in ft) is *h* = 5280 tan(𝜃).

From the figure, we see that

tan(𝜃) =

*h*

1

mi

,

so

*h*

=

(tan(𝜃))

1

mi

=

(tan(𝜃))

1

mi

5280

ft

1 mi

=

ft.
(b) Complete the table to find the height of the rocket at the given angles of elevation. (Round your answers to the nearest foot.)

**𝜃**

40°

60°

65°

70°

***h***

### (a) Deriving the Height Formula

Given:
- A rocket is fired vertically and is observed by an observer 1 mile away from the rocket's initial launch point.
- The angle of elevation (𝜃) from the observer's point to the rocket is related to the height of the rocket.

The key relationship here is given by the **tangent of the angle of elevation**:

\[
\tan(𝜃) = \frac{h}{d}
\]
where:
- \( h \) is the height of the rocket (in feet),
- \( d \) is the horizontal distance from the observer to the rocket (in miles), which is 1 mile in this case.

Since the distance from the observer to the rocket is **1 mile**, and 1 mile is equivalent to 5280 feet, we substitute:

\[
\tan(𝜃) = \frac{h}{5280} \quad \text{(since 1 mile = 5280 ft)}
\]

Solving for \( h \):

\[
h = 5280 \cdot \tan(𝜃)
\]

Thus, the height \( h \) of the rocket (in feet) is related to the angle of elevation \( \theta \) by:

\[
h = 5280 \cdot \tan(𝜃)
\]

### (b) Completing the Table

Now, we will calculate the height of the rocket for each given angle of elevation (rounded to the nearest foot):

Given angles:
- \( \theta = 40^\circ \)
- \( \theta = 60^\circ \)
- \( \theta = 65^\circ \)
- \( \theta = 70^\circ \)

We use the formula \( h = 5280 \cdot \tan(𝜃) \).

Let me calculate these values for you.Here is the completed table for the height of the rocket at the given angles of elevation (rounded to the nearest foot):

\[
\begin{array}{|c|c|}
\hline
\text{Angle of Elevation (𝜃)} & \text{Height of the Rocket (h) in feet} \\
\hline
40^\circ & 4430 \\
60^\circ & 9145 \\
65^\circ & 11323 \\
70^\circ & 14507 \\
\hline
\end{array}
\]

Would you like further details or explanations? Feel free to ask!

Here are 5 related questions you could explore:
1. What happens to the height as the angle approaches 90°?
2. How would the formula change if the observer were farther than 1 mile?
3. How does the rate of change of height vary with the angle?
4. Can we find the velocity of the rocket if we know the change in angle over time?
5. How does the tangent function behave at different angles in terms of rocket height?

**Tip:** As the angle of elevation increases, the tangent function increases rapidly, causing the height to grow exponentially.

A rocket fired straight up is tracked by an observer on the ground 1 mi away. (a) Show that when the angle of elevation is 𝜃, the height of the rocket (in ft) is *h* = 5280 tan(𝜃). (b) Complete the table to find the height of the rocket at the given angles of elevation. (Round your answers to the nearest foot.) θ = 40°, 60°, 65°, 70°.

Learn how to calculate the height of a rocket using trigonometric functions and the angle of elevation. This problem demonstrates the relationship between the tangent function and right triangles to find the height at various angles of elevation. Complete with a table for angles 40°, 60°, 65°, and 70°.

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