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Math Help - Word problems...

  1. #1
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    Word problems...

    Two side of parallelogram form a 120 angle. The longer side of the parallelogram is 18cm long and the diagonal opposite the 120 angle is 24cm long. Find the length of the shorter side and the are of the parallelogram?

    Somebody help me? please?
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  2. #2
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    Start by using the sine rule.
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  3. #3
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    i can't start >.< i don't know the diagram or the sketch of the problem i'm confused>.<
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  4. #4
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    The diagonal opposite goes from the top left corner to the bottom right corner.
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  5. #5
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    Hello, cutiepie32!

    Did you make a sketch?


    Two side of parallelogram form a 120 angle.
    The longer side of the parallelogram is 18cm long.
    The diagonal opposite the 120 angle is 24cm long.
    Find: (a) the length of the shorter side
    . e . .(b) the area of the parallelogram
    Code:
        D o - - - - - - - o C
          :\ *             \
          : \ θ *           \
          :  \     *  24     \
          h   \ x     *       \
          :    \         *     \
          :     \           *   \
          :      \ 120      α * \
          o     A o - - - - - - - o B
          E              18

    We have parallelogram ABCD with: . \angle A = 120^o,\;\;AB = 18,\;\;BD = 24
    Let: . \theta = \angle BDA,\;\;\alpha = \angle DBA,\;\;x = DA


    Law of Sines: . \frac{\sin\theta}{18} \:=\:\frac{\sin120^o}{24} \quad\Rightarrow\quad \sin\theta \:=\:0;541265877

    . . Hence: . \theta \:\approx\:32.77^o

    . . Then: . \alpha \;=\;180^o - 120^o - 32.77^o \;=\;27.23^o


    Law of Sines: . \frac{x}{\sin27.23^o} \:=\:\frac{24}{\sin120^o} \quad\Rightarrow\quad x \:\approx\:12.68\text{ cm} .(a)



    In right triangle DEB\!:\;\sin\alpha \,=\,\frac{h}{24} \quad\Rightarrow\quad h \;=\;24\sin27.23^o \:\approx\:10.98


    Therefore: . \text{Area} \;=\;bh \;=\;(18)(10.98) \;=\; 197.64\text{ cm}^2 .(b)

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  6. #6
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    uhm thx but where did you get the sin=0;541265877?
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