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Math Help - [SOLVED] Urgentt HELP! 3D Trigonometry Problem

  1. #1
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    Arrow [SOLVED] Urgentt HELP! 3D Trigonometry Problem

    Hii I reallyy need to solve this question. could you guys help me and also solve it? thankss

    Two buildings of equal height are 40m apart and at a point on the horizontal line joining their feet the angles of elevation of the top of the building are 47 degrees and 28 degrees.

    Show that the height (h) of the building is given by:

    *=Degrees

    h = 40tan47*tan28*
    ----------------
    tan47*+tan28*

    ----------=Divider/Over

    Show all working.

    Thanks guys
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  2. #2
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    Hi



    \tan \theta_1 = \frac{h}{d+d'} [1]

    \tan \theta_2 = \frac{h}{d'} [2]

    Replace d' in [1] by its expression in terms of h and \tan \theta_2 derived from [2]

    EDIT : the point seems to be actually BETWEEN the 2 buildings
    That does not change the method anyway

    Last edited by running-gag; March 7th 2009 at 01:42 AM.
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  3. #3
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    Heyy Uhh

    Could you explain it a bit more..

    Im confused a bit..

    and could you also solve it?

    also what does it have to do with the h = 40tan47*tan28* thing? im kinda dumb. please help out

    Thanks
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  4. #4
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    Take the second scheme of my previous post

    d = 40 m
    \theta_1 = 28 deg
    \theta_2 = 47 deg

    The tangent is defined as the ratio of the opposite leg to the adjacent leg

    \tan \theta_1 = \frac{h}{d-d'} [1]

    \tan \theta_2 = \frac{h}{d'} [2]

    Replace d' = \frac{h}{\tan \theta_2} in [1] to find the expression of h with respect to d, \tan \theta_1 and \tan \theta_2
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  5. #5
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    Heyy

    I got up to here..
    what to I do after this?

    /tan/theta_1 = \frac{h}{d-\frac{h}{/tan/theta_2}}

    basically

    tan theta 1 = h over d minus h over tan theta 2
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  6. #6
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    OK so you are getting

    \tan \theta_1 = \frac{h}{d-\frac{h}{\tan \theta_2}} = \frac{h \:\tan \theta_2}{d \:\tan \theta_2-h}

    d \:\tan \theta_2-h) = h \:\tan \theta_2" alt="\tan \theta_1\d \:\tan \theta_2-h) = h \:\tan \theta_2" />

    d\:\tan \theta_1 \: \tan \theta_2 = h \:\tan \theta_2 + h \:\tan \theta_1

    Finally

    h = \frac{d\:\tan \theta_1 \: \tan \theta_2}{\tan \theta_1 + \tan \theta_2}
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  7. #7
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    Wow..

    Thanks alot

    I get nearly everything

    cept for the second last part.

    what happens to the 2 h's? doesnt it get like added up or something?

    explain please.

    Thanks
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  8. #8
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    \tan \theta_1 = \frac{h}{d-\frac{h}{\tan \theta_2}}

    Multiplying numerator and denominator by \tan \theta_2

    \tan \theta_1 = \frac{h \:\tan \theta_2}{d \:\tan \theta_2-h}

    Multiplying both sides by d \:\tan \theta_2-h

    d \:\tan \theta_2-h) = h \:\tan \theta_2" alt="\tan \theta_1\d \:\tan \theta_2-h) = h \:\tan \theta_2" />

    Expanding

    d\:\tan \theta_1 \: \tan \theta_2 - h \:\tan \theta_1 = h \:\tan \theta_2

    Adding h \:\tan \theta_1 to both sides

    d\:\tan \theta_1 \: \tan \theta_2 = h \:\tan \theta_2 + h \:\tan \theta_1

    Factorizing the RHS

    \tan \theta_1 + \tan \theta_2)" alt="d\:\tan \theta_1 \: \tan \theta_2 = h\\tan \theta_1 + \tan \theta_2)" />

    Dividing both sides by \tan \theta_1 + \tan \theta_2

    h = \frac{d\:\tan \theta_1 \: \tan \theta_2}{\tan \theta_1 + \tan \theta_2}
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  9. #9
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    Ooof.

    Wow.

    Thanks alot.

    I get it now! Geez I was dumb

    Thanks,

    You Rock xD

    SOLVED
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