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Math Help - [SOLVED] Few beg. Trig Probs Angles

  1. #1
    Junior Member
    Joined
    Aug 2008
    Posts
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    [SOLVED] Few beg. Trig Probs Angles

    So when you have 90-x in a right triangle that you are trying to solve cos sin and tang etc.. those questions, how do you include angle degrees into your calculations.. For instance,

    l there is a h(3) line
    4 l i just can't
    l draw online
    l_____________x
    5

    I know how to simply find these functions by dividing one another but I get confused when it comes to including angles.


    I also have a question regarding "clinometer" use.

    Well, we can use such an instrument (called a “clinometer”) to measure the angle A. Then we know that the angle between the ground and the building is a right angle, so we know the measure of 3 angles, because they all add up to 180. Now all we have to do is find a similar triangle anywhere. Then we can measure the ratio of the sides that correspond to the height and the ground length, and then we can multiply by the actual ground length to get the height.

    So once you have the three angles, how do you find the measurement on the structure you are trying to identify. I am given a question like this:

    Given the angle you are given to the top of the building, calculate the height given distance?

    I am unsure how to find the height of the building could someone help?

    Here is a sample prob I am given,

    Angle 71 Distance 20 Meters

    Top

    L
    l Angle
    ; in
    l box
    l
    _____________________________71
    20
    Any potential help appreciated.
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  2. #2
    No one in Particular VonNemo19's Avatar
    Joined
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    From
    Detroit, MI
    Posts
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    Quote Originally Posted by KevinVM20 View Post
    I am unsure how to find the height of the building could someone help?

    Here is a sample prob I am given,

    Angle 71 Distance 20 Meters

    Top

    L
    l Angle
    ; in
    l box
    l
    _____________________________71
    20
    Any potential help appreciated.
    Since tan{\theta}=\frac{o}{a}, let \theta=71^\circ, and let a=20, and let o= the side opposite \theta (which is the height of the building).

    \tan{71^\circ}=\frac{o}{20}

    Therefore,

    (20)\tan{71^\circ}=o

    Does that help?

    Note that in this problem we didn't have to find any of the other angles to find the height of the building.
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