Question

From the top of a tower 50 meters high, the angle of depression to a point on the ground is 30°. What is the distance from the base of the tower to that point on the ground? (Assume the ground is horizontal)

• A. \(\frac{25}{3}\) meters
• B. \(\frac{50}{3}\) meters
• C. \(\frac{50}{\sqrt{3}}\) meters
• D. 100 meters

Hint:

Use the tangent function for problems involving height and horizontal distance, especially when the angle of depression or elevation is given.

Answer 1

The Correct Option is B

Let the distance from the base of the tower to the point on the ground be \(d\). Given: - Height of the tower = 50 meters, - Angle of depression = 30°. We know that the angle of depression is equal to the angle of elevation from the point on the ground to the top of the tower. Using the tangent function in the right triangle formed by the height of the tower, the horizontal distance, and the line of sight: \[ \tan(30^\circ) = \frac{\text{Height of the tower}}{\text{Distance from the base}} = \frac{50}{d} \] We know that \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\), so: \[ \frac{1}{\sqrt{3}} = \frac{50}{d} \] Solving for \(d\): \[ d = 50 \times \sqrt{3} = \frac{50}{\sqrt{3}} = \frac{50 \times \sqrt{3}}{3} = \frac{50}{3} \] Thus, the distance from the base of the tower to the point on the ground is \(\frac{50}{3}\) meters.
Final answer
Answer: \(\boxed{\frac{50}{3}}\)