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Math Help - angles of elevation and depression

  1. #1
    salahu31
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    angles of elevation and depression

    Could you please help me with this problem thnx.From a coastal lookout point A,80m above sea level,a man sights two boats B and C in the same direction.The angles of depression of the two boats are 13degrees and 24degrees respectively.Find the distances of B and C from the point below A and so find find the distance between the boats.thanks againsalahu
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  2. #2
    Super Member

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    Hello, salahu!

    From a coastal lookout point A, 80m above sea level,
    a man sights two boats B and C in the same direction.
    he angles of depression of the two boats are 13 and 24. respectively.
    Find the distances of B and C from the point below A
    and so find find the distance between the boats.
    Code:
        A * - - - - - - - - - - - - - - E
          |  *  *  13
          |     *     *
       80 |        *        *
          |           *           *
          |          24 *         13  *
          * - - - - - - - - * - - - - - - - - *
          D       x         C       y         B

    \angle EAB = 13^o.\;,\text{ Hence, }\angle ABD = 13^o

    \angle EAC = 24^o.\;\text{ Hence, }\angle ACD = 24^o

    Let x \:=\:DC.\;y \:=\:CB


    In right triangle ADC\!:\;\;\tan24^o \;=\;\frac{80}{x}\quad\Rightarrow\quad x \;=\;\frac{80}{\tan24^o} .[1]

    In right triangle ADC\!:\;\;\tan13^o \;=\;\frac{80}{x+y}\quad\Rightarrow\quad x + y \;=\;\frac{80}{\tan13^o} .[2]

    Substitute [1] into [2]: . \frac{80}{\tan24^o} + y \;=\;\frac{80}{\tan24^o}

    Therefore: . y \;=\;\frac{80}{\tan13^o} - \frac{80}{\tan24^o} \;=\;166.835128 m.

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