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Thread: sine rule

  1. #1
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    sine rule

    In triangle ABC , AB = x cm , BC = (4-x) cm , angle BAC = y degrees and angle BCA = 30 degrees.
    Given that sin y = \frac{1}{\sqrt{2}}, show that x=4(\sqrt{2}-1)

    what i got is:

    \frac{(4-x)}{\sin y}=\frac{x}{\sin 30}\\x=2\sqrt{2}-\frac{\sqrt{2}x}{2}\\2x=4\sqrt{2}-\sqrt{2}x\\x(2-\sqrt{2})=4\sqrt{2}\\x=2\sqrt{2}-4\\x=2(\sqrt{2}-2)

    where did i go wrong ?
    Last edited by AbYz; Aug 7th 2017 at 11:12 AM.
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  2. #2
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    Re: sine rule

    Law of sines ...

    $\dfrac{x}{\sin(30)} = \dfrac{4-x}{\sin{y}}$

    $2x = \sqrt{2}(4-x)$

    $2x = 4\sqrt{2} - x\sqrt{2}$

    $2x + x\sqrt{2} = 4\sqrt{2}$

    $x(2 + \sqrt{2}) = 4\sqrt{2}$

    $x = \dfrac{4\sqrt{2}}{2+\sqrt{2}} \cdot \dfrac{2-\sqrt{2}}{2-\sqrt{2}} = \dfrac{8\sqrt{2}-8}{2} = 4\sqrt{2} - 4 = 4(\sqrt{2}-1)$
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  3. #3
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    Re: sine rule

    Quote Originally Posted by AbYz View Post
    In triangle ABC , AB = x cm , BC = (4-x) cm , angle BAC = y degrees and angle BCA = 30 degrees.
    Given that sin y = \frac{1}{\sqrt{2}}, show that x=4(\sqrt{2}-1)

    what i got is:

    \frac{(4-x)}{\siny}=\frac{x}{\sin30}
    x=2\sqrt{2}\-\frac{sqrt{2}x}{2} x=2\sqrt{2}-4 x=2(\sqrt{2}-2)

    where did i go wrong ?
    sin(30)= 1/2, not \sqrt{2}/2
    Last edited by skeeter; Aug 7th 2017 at 01:04 PM. Reason: fix typo for sin(30)
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