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Trig

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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 oppisite side of the runway are 6 degrees and 13 degreess respectivel. The points and the tower are in the same vertical plane and the distance from A to B is 1.1km. Determine the height of the tower. Please give me a diagram and step by step instructions. This question has me stumped!!!

  • Trig - ,

    Let the base of the tower be C, and the top of the tower be T. So, the height h=CT, and we have

    AB=1, so AC=x and BC=1-x

    h/x = tan6°
    h/(1-x) = tan13°

    so, solve for x:

    xtan6° = (1-x)tan13°
    x = 0.687

    h = 0.0722km = 72.2m

  • Trig - ,

    Can you show me the steps you used to isolate x?

  • Trig - ,

    x is the distance from A to the base of the tower.
    Since AB=1, the distance to B is 1-x.

    Did you not understand the diagram points?

  • Trig -- oops - ,

    oops. Sorry. I see that AB=1.1
    So, adjust the equations and solve again.

  • Trig - ,

    Yes I understand the diagram points and can see how you came up with this now but am having trouble coming up with an answer for
    xtan6°=(1.1-x)tan13°
    I know this is basic but I am having trouble isolating the x!
    I am getting
    x=tan13°-tan6°/1.1 which is giving me a negative answer.

  • Trig - ,

    plugging in the trig values, you have
    .1051x = .2309(1.1-x)
    Have you forgotten your algebra I?
    .1051x = .2540 - .2309x
    .3360x = .2540
    x = 0.756

  • Trig - ,

    Lol! Apparently I have been working on this question too long! Thanks!

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