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-   -   Trigonometry- height & length apart (https://www.askmehelpdesk.com/showthread.php?t=230250)

  • Jun 24, 2008, 09:57 AM
    Umpalumpa
    Trigonometry- height & length apart
    I really don't get this:S
    A video camera is mounted at the top of a 120m building. When the camera tilts down 36(degrees) with the horizontal, it views the bottom of another building. If it tilts up at 47 (degrees) to the horizontal, it can view the top of the same building.
    a) How far apart are the two buildings?
    b) how tall is the building that the camera sees?

    HELP!:-|:confused: :(
  • Jun 24, 2008, 10:47 AM
    galactus
    1 Attachment(s)
    Draw a picture. That always helps. Use the law of tangents to find x and y.

    Sorry, my diagram is not too scale but it gives the idea.
  • Jun 24, 2008, 12:02 PM
    brandizzle
    your formulas would be tan36=x/120 for how far apart the buildings are. Then plug in x for the formula for the height of second building--- tan47=x/y.
  • Jun 25, 2008, 04:14 AM
    Unknown008
    Hum... brandizzle? You messed up woth your formulas. The formula is tan36=120/x instead of tan36=x/120 and tan47=y/x instead of tan47=x/y...

    And you should then add y to 120 to get the height of the building, not just stopping at tan47=y/x.

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