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Math Help - length of pole

  1. #1
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    length of pole

    Im suppose to find how high a building is [ not literally, lol ].

    A flagpole 40ft stands on top of a building. The angle of elavation for the top of the pole is: 54 degrees 54' and the angle of elevation for the bottom of the pole is 47 degrees 30'. How high is the building?

    I did:

    tan 54 54' = 40
    x

    x= 28.11

    and then did tan 47 30' = 40
    x

    x= 36.65

    and subtract that and get a difference of 8.54.

    Is that how you find the height of the building?
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  2. #2
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    \tan(54^{\circ}54') = \frac{40+h}{x}

    \tan(47^{\circ}30') = \frac{h}{x}

    where x is the unknown horizontal distance to the building and h is the height of the building.

    using the 2nd equation ...

    x = \frac{h}{\tan(47^{\circ}30')}

    substitute for x in the 1st equation and solve for h.
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  3. #3
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    how do you go about solving for h?
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  4. #4
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    \tan(54^{\circ}54') = \frac{40+h}{x}

    x = \frac{h}{\tan(47^{\circ}30')}

    substitute for x in the 1st equation and solve for h ...

    \tan(54^{\circ}54') = \frac{(40+h)\tan(47^{\circ}30')}{h}

    h\tan(54^{\circ}54') = (40+h)\tan(47^{\circ}30')

    h\tan(54^{\circ}54') = 40\tan(47^{\circ}30') + h\tan(47^{\circ}30')

    h\tan(54^{\circ}54') - htan(47^{\circ}30') = 40\tan(47^{\circ}30')

    h[\tan(54^{\circ}54') - tan(47^{\circ}30')] = 40\tan(47^{\circ}30')

    h = \frac{40\tan(47^{\circ}30')}{\tan(54^{\circ}54') - tan(47^{\circ}30')}
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