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Math Help - find angle

  1. #1
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    find angle

    A cable to the top of a tower makes an angle of 35 deg. with the level ground. At a point 100 yards closer to the tower the angle of elevation to the top of the tower is 59 deg. Estimate the length of the cable.

    Please help
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  2. #2
    Bar0n janvdl's Avatar
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    Quote Originally Posted by ccdalamp View Post
    A cable to the top of a tower makes an angle of 35 deg. with the level ground. At a point 100 yards closer to the tower the angle of elevation to the top of the tower is 59 deg. Estimate the length of the cable.

    Please help
    See the attachment.

    c represents the cable.
    t represents the tower.


    From the tower to point A is x metres.
    From the tower to point B is (x - 100) metres.

    So the length of the area between points A and B is 100 metres.

     \frac{c}{sin 121} = \frac{100}{sin 24}

     c = \frac{100 \ Sin 121}{sin 24}

     c = 210,74 \ metres

    Yes the picture is ugly, but you get the idea!
    Attached Thumbnails Attached Thumbnails find angle-cable.jpg  
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  3. #3
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    Hello, ccdalamp!

    This type of problem can be solved without the Law of Sines,
    . . but it requires more work.


    A cable to the top of a tower makes an angle of 35 with the level ground.
    At a point 100 yards closer to the tower, the angle of elevation to the top
    of the tower is 59. .Estimate the length of the cable.
    Code:
                                      * D
                                  * * |
                              *   *   |
                      L   *     *     |
                      *       *       |t
                  *         *         |
              * 35       * 59       |
          * - - - - - - * - - - - - - *
          A     100     B      x      C

    In the diagram: . AB = 100,\:BC = x,\:CD = t,\:AD = L

    In right triangle DCB\!:\;\tan59^o \,= \,\frac{t}{x}\quad\Rightarrow\quad t \,= \,x\tan59^o . [1]
    In right triangle DCA\!:\;\tan35 \,=\,\frac{t}{x+100}\quad\Rightarrow\quad t \,= \,(x+100)\tan35 . [2]

    Equate [1] and [2]: . x\tan59^o \:=\:(x+100)\tan35^o

    . . Solve for x\!:\;\;x \;=\;\frac{100\tan35^o}{\tan59^o-\tan35^o} \;\approx\;72.63


    In right triangle DCA\!:\;\cos35^o \:=\:\frac{x+100}{L}\quad\Rightarrow\quad L \:=\:\frac{x+100}{\cos35^o}

    Therefore: . L \;=\;\frac{72.63 + 100}{\cos35^o} \;\approx\;210.74 yards.


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