A tower subtends an angle of 30º at a point on the same level as the foot of the tower. At a second point h meters above the first, the depression of the foot of the tower is 60º. What is the horizontal distance of the tower from the point?
Step-by-step Solution:
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.
Let's define the points from the problem description:
The points F, P1, and P2 form a right-angled triangle ($\triangle FP_1P_2$), with the right angle at P1.
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 $\angle FP_1 = 60^\circ$.
Now we can use a trigonometric ratio in $\triangle FP_1P_2$ to relate the angle $60^\circ$ to the sides $h$ and $x$.
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.)