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Thread: Trig Question

  1. #1
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    Trig Question



    can someone solve the triange im really stuck on it
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  2. #2
    Senior Member tukeywilliams's Avatar
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    use law of cosines/ law of sines.
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  3. #3
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    well i know that since the chapter is based on it i just dont kno where to start
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  4. #4
    Senior Member tukeywilliams's Avatar
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    Is that $\displaystyle <AXB $?
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  5. #5
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    ye my bad its <axb
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  6. #6
    Senior Member tukeywilliams's Avatar
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    So $\displaystyle \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} $.

    $\displaystyle \frac{50}{\sin 81.65} = \frac{b}{\sin 60} $. Solve for $\displaystyle b $. Then you can solve for the height from this information. So $\displaystyle \frac{h}{\sin 42.6} = \frac{b}{\sin 47.4} $. We already know $\displaystyle b $. So solve for $\displaystyle h $. So $\displaystyle h = \frac{b \sin 42.6}{\sin 47.4} $ or $\displaystyle h \approx 3.60b = 1748.661 $.
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  7. #7
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    i get everythin but how du u know what angles to use

    like if its <XAB which angle is it referrin too

    a rite
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  8. #8
    Senior Member tukeywilliams's Avatar
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    The angles refer to the opposite sides. So $\displaystyle <XAB $ refers to side $\displaystyle BX $.
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