- 0. Functions7h 52m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms34m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation3h 18m
- 4. Applications of Derivatives2h 38m
- 5. Graphical Applications of Derivatives6h 2m
- 6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
- 7. Antiderivatives & Indefinite Integrals1h 26m
- 8. Definite Integrals3h 25m
0. Functions
Introduction to Trigonometric Functions
Problem 1.4
Textbook Question
A hot-air balloon rising straight up from a level field is tracked by a range finder located 500 ft from the point of liftoff. Express the balloon’s height as a function of the angle the line from the range finder to the balloon makes with the ground.

1
Identify the right triangle formed by the balloon, the range finder, and the point on the ground directly below the balloon. The balloon's height is the opposite side, the distance from the range finder to the point of liftoff is the adjacent side, and the line from the range finder to the balloon is the hypotenuse.
Let \( \theta \) be the angle between the ground and the line from the range finder to the balloon. The height of the balloon, \( h \), is the side opposite to \( \theta \).
Use the tangent function, which relates the opposite side to the adjacent side in a right triangle: \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{500} \).
Solve for the height \( h \) in terms of \( \theta \): \( h = 500 \cdot \tan(\theta) \).
Thus, the balloon's height as a function of the angle \( \theta \) is \( h(\theta) = 500 \cdot \tan(\theta) \).
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