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Math Help - Trig

  1. #1
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    Trig

    A simple question just cant work out which function to use.

    A man is on the top of the cliff 125m high. He observes two boats in line with him at angles of depression of 45 degrees and 75degrees. How far are they from each other.

    Any help would be great,

    Thank You
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  2. #2
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    Quote Originally Posted by n3_olp View Post
    A simple question just cant work out which function to use.

    A man is on the top of the cliff 125m high. He observes two boats in line with him at angles of depression of 45 degrees and 75degrees. How far are they from each other.

    Any help would be great,

    Thank You
    We are told the height and the angle of depression which the angle down from the horizontal. We have to find the angle between the depression and the height so that we can work out opposite given the angle and the adjacent height.

    For the 45^{\circ} boat:

    Angle between the depression and the height: 90^{\circ} - 45^{\circ} = 45^{\circ}

    \text{tan}\theta = \frac{\text{Opposite}}{\text{Adjacent}} \implies\text{Opposite} = (\text{tan}\theta ) \times  {\text{Adjacent}}

    \text{Opposite} = (\text{tan}\theta ) \times  {\text{Adjacent}} = (\tan(45^{\circ}))(125) = 125m

    For the 75^{\circ} boat:

    Angle between the depression and the height: 90^{\circ} - 75^{\circ} = 15^{\circ}

    \text{Opposite} = (\text{tan}\theta ) \times  {\text{Adjacent}} = (\tan(1 5^{\circ}))(125) \approx 33.5m

    Therefore the distance between each boat is 125m - 33.5m = 91.5m
    Attached Thumbnails Attached Thumbnails Trig-diagram.jpg  
    Last edited by Simplicity; June 30th 2008 at 01:00 AM. Reason: Attaching Badly Drawn Diagram
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