Post a New Question


posted by .

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.

  • Math -

    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

  • Math -

    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 This Question

First Name:
School Subject:

Related Questions

More Related Questions

Post a New Question