Question

from an observation tower that overlooks a runway, the angles of depression of point a, on one side of the runway, and point b, on the opposite side of the runway are 6 degrees and 13 degrees, respectively. the points and the tower are in the same vertical plane and the distance from a to b is 1.1 km. whats the height of the tower

Answer 1

Let the tower height be h. Let the horizontal distances from the tower base to A and B be x_A and x_B (on opposite sides). From the angles of depression,

tan 6° = h / x_A => x_A = h / tan 6°
tan 13° = h / x_B => x_B = h / tan 13°

The distance A to B is x_A + x_B = 1.1 km, so
h(1/tan6° + 1/tan13°) = 1.1 km.

Thus
h = 1.1 / (1/tan6° + 1/tan13°).

Using tan6° ≈ 0.105104 and tan13° ≈ 0.230868 gives
h ≈ 1.1 / (9.51436 + 4.33013) ≈ 1.1 / 13.84449 ≈ 0.07947 km ≈ 79.5 m.

So the tower is about 79.5 meters tall (≈ 79 m).