To solve this problem, we can use trigonometry. Specifically, the tangent function is helpful in right triangles, where the angle of depression forms one of the angles.

Problem Breakdown:

• The angle of depression is 40°.
• The horizontal distance to the landmark is 799 feet.
• We are looking for the vertical distance (height) of the balloon, which is the opposite side in the triangle.

Trigonometric Setup:

The tangent of an angle in a right triangle is given by:

$\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$

Where:

• $\theta = 40^\circ$ is the angle of depression.
• The opposite side is the vertical height of the balloon (which we're solving for).
• The adjacent side is the horizontal distance to the landmark, which is 799 feet.

Step-by-step solution:

$\tan(40^\circ) = \frac{h}{799}$

Where $h$ is the vertical height of the balloon. Rearranging the equation to solve for $h$:

$h = 799 \times \tan(40^\circ)$

Now we calculate:

$h = 799 \times 0.8391 \quad (\text{using the value of} \, \tan(40^\circ) \approx 0.8391)$

$h \approx 670.72 \, \text{feet}$

So, the balloon's vertical distance above the ground is approximately 670.72 feet.

Final Answer:

The balloon is about 670.72 feet above the ground.

Would you like any further details or have any questions about this calculation? Here's a set of related questions for practice:

1. How would the result change if the angle of depression were 30° instead of 40°?
2. If the horizontal distance were increased to 1,000 feet, what would the vertical distance be at the same angle of depression?
3. How would you solve for the horizontal distance if the vertical height was known instead?
4. What are other trigonometric ratios that can be used to solve for different sides of a right triangle?
5. How do you convert degrees to radians, and how would this affect trigonometric calculations?

Tip: When dealing with angles of depression or elevation, always ensure to set up your triangle correctly and note whether the angle is measured from the horizontal or vertical.