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. Determine the height of the tower.

Can someone help me solve this? I am confused on how this is laid out, can someone draw some kind of picture, i have no idea where to put the angles and tower/road.

Answer 1

Assume we are taking a side view,
draw a straight line, label the left end A and the right end B
Place a point between them to represent the tower, and give it some height.
Join both A and B to the top of the tower
You now have 2 right-angled triangles.
where the base angle at A is 6° and at B is 13°

let the distance from A to the base of the tower be x
then the distance from B to the base of the tower is 1.1 - x
let the height of the tower be h
in the triangle containing A
tan 6° = h/x ---> h = xtan6
in the triangle containing B
tan 13° = h/(1.1-x) ---> h = (1.1 - x)tan13

xtan6 = 1.1tan13 - xtan13
x(tan6 + tan13) = 1.1tan13
x = 1.1tan13/(tan6 + tan13) = .3441195

h = xtan6 = .03616..
all units are in km, so the tower is .036 km or appr 36 m high

Answer 2

thank you, makes more sense, i thought it was supposed to look like that but its just worded poorly, and the vertical plane got me confused.

Answer 3

Calculation error on 1.1tan13/(tan6 + tan13), should be approx. 0.7559.