The trail, the ground, and the vertical rise form a right triangle. The trail length, 200 feet, is the hypotenuse, and the height above the parking area is opposite the $30^{\circ }$ angle. Use sine: $\sin (30^{\circ })=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{h}{200}$. Since $\sin (30^{\circ })=\frac{1}{2}$, we have $\frac{1}{2}=\frac{h}{200}$. Multiply both sides by 200: $h=100$. So the lookout point is about 100 feet above the parking area.

The triangle has vertices $(0,0)$, $(4,0)$, and $(4,3)$. For angle $\theta$ at the origin, the horizontal leg from $(0,0)$ to $(4,0)$ has length $4$, and the vertical leg from $(4,0)$ to $(4,3)$ has length $3$. Tangent is the ratio of opposite side to adjacent side, so

Therefore, the correct answer is $\frac{3}{4}$.

Segment $PQ$ is horizontal from $(0,0)$ to $(8,0)$, so its length is $8$. Segment $QR$ is vertical from $(8,0)$ to $(8,6)$, so its length is $6$. This makes a right triangle with right angle at $Q$. For $\angle QPR$, the opposite side is $QR=6$ and the adjacent side is $PQ=8$. Therefore,

The line passes through $(0,0)$ and $(4,3)$, so the right triangle has horizontal leg length $4$ and vertical leg length $3$. For the acute angle $\theta$ between the line and the $x$-axis,

Therefore, the correct answer is $\frac{3}{4}$.

Use the first row to find the drone's height in terms of the given angle. For the first position, $\tan 35^{\circ }=\frac{\text{height}}{120}$, so the height is $120\tan 35^{\circ }$. Since the drone stays at a constant height, for the second position, $\tan \theta =\frac{\text{height}}{160}=\frac{120\tan 35^{\circ }}{160}$. Therefore, the correct answer is $\frac{120\tan 35^{\circ }}{160}$.