Correct Answer: B ($h \cot 60^\circ$)

This heights and distances problem can be solved by focusing on the right-angled triangle formed by the two observation points and the foot of the tower.

Step 1: Set up the Geometry

Let's define the points from the problem description:

• Let F be the foot of the tower on the ground.
• Let P1 be the first observation point, on the same level as F.
• Let P2 be the second observation point, located at a height h vertically above P1.
• The horizontal distance we need to find is FP1. Let's call this distance x.

The points F, P1, and P2 form a right-angled triangle ($\triangle FP_1P_2$), with the right angle at P1.

Step 2: Use the Angle of Depression

The problem states that the angle of depression of the foot of the tower (F) from the second point (P2) is $60^\circ$.

The angle of depression from P2 to F is the angle between the horizontal line from P2 and the line segment $P_2F$. Due to the property of alternate interior angles, this is equal to the angle of elevation of P2 from F.

Therefore, in our right-angled triangle $\triangle FP_1P_2$, the angle at the foot of the tower is $\angle FP_1 = 60^\circ$.

Step 3: Apply Trigonometry to Find the Distance

Now we can use a trigonometric ratio in $\triangle FP_1P_2$ to relate the angle $60^\circ$ to the sides $h$ and $x$.

• The side opposite to the $60^\circ$ angle is $P_1P_2$, which has a length of $h$.
• The side adjacent to the $60^\circ$ angle is $FP_1$, which has a length of $x$.

Using the cotangent ratio ($\cot\theta = \frac{\text{Adjacent}}{\text{Opposite}}$):

$$ \cot(60^\circ) = \frac{FP_1}{P_1P_2} = \frac{x}{h} $$

Solving for the horizontal distance $x$ gives:

$$ x = h \cot 60^\circ $$

(Note: The information about the $30^\circ$ angle of elevation to the top of the tower is extra information not needed to solve for the horizontal distance.)