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Math Help - final question on bearings

  1. #1
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    final question on bearings

    The angle of elevation from the top of the building 200 feet high to the top of a taller building 100 yards away is 10 degree. Find the height of the taller building correct to the nearest tenth of a foot.

    100yards=300ft


    200 is the short building. 300 is the adjacent
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cocoknny View Post
    The angle of elevation from the top of the building 200 feet high to the top of a taller building 100 yards away is 10 degree. Find the height of the taller building correct to the nearest tenth of a foot.

    100yards=300ft


    200 is the short building. 300 is the adjacent
    see the diagram below. the black rectangle is the short building, the gray rectangle is the tall building.

    they are 300 feet apart. and the angle of elevation is 10 degrees. thus we can use the tangent ratio to find BC.

    Now, \tan \left( 10^o \right) = \frac {BC}{300}

    \Rightarrow BC = 300 \tan \left( 10^o \right)

    So, the height, h, of the taller building is given by:

    h = 200 + BC

    i suppose you can take it from here
    Attached Thumbnails Attached Thumbnails final question on bearings-building.gif  
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  3. #3
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    hello

    can u explain why 200ft is added even though its the shorter building?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cocoknny View Post
    can u explain why 200ft is added even though its the shorter building?
    look at the diagram. the taller building is taller by the length BC. so its height is the height of the short building plus BC. so it is 200 + BC
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