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    There’s this question that I’m stuck on- I don’t know where to go with it at all:
    In the triangle ABC, AB=x cm, BC=(4-x) cm, angle BAC=y° and angle BCA= 30°. Given that siny°=1/square root 2 show that x=4(square root 2 - 1)
    I’ve tried doing the sine rule and switching the numbers and all about to end up there but somehow it hasn’t worked. I’d really appreciate the help, thank you
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    So we have the sin rule:
    SinA/a = SinB/b

    Using what we have been told about the triangle:
    sinY/(4-x) = sin30/(x)
    (1/√2)/(4-x) = 0.5/x <--- sin30 = 1/2

    Re-arrange the equation to get x, resulting in x = 4(√2 - 1)

    If you need additional help rearranging, I can show you some more steps.
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    (Original post by FryOfTheMann)
    So we have the sin rule:
    SinA/a = SinB/b

    Using what we have been told about the triangle:
    sinY/(4-x) = sin30/(x)
    (1/√2)/(4-x) = 0.5/x <--- sin30 = 1/2

    Re-arrange the equation to get x, resulting in x = 4(√2 - 1)

    If you need additional help rearranging, I can show you some more steps.
    Oh my god this is incredibly helpful, thank you! Would you mind showing me further instructions on how to rearrange it, though? I’ve gotten (1/root2)/(4-x) x 0.5= x but I’m not even sure if that’s right or where to go from there :/
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    (Original post by geometric)
    Oh my god this is incredibly helpful, thank you! Would you mind showing me further instructions on how to rearrange it, though? I’ve gotten (1/root2)/(4-x) x 0.5= x but I’m not even sure if that’s right or where to go from there :/
    Straight from the information you know (siny)/(4-x) = (sin30)/x
    ====> (1/root2)/(4-x) = 1/2x
    ====> 2x/root2 = 4-x

    proceed to solve from there (Hint: you will need to rationalise denominators)
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    (Original post by geometric)
    Oh my god this is incredibly helpful, thank you! Would you mind showing me further instructions on how to rearrange it, though? I’ve gotten (1/root2)/(4-x) x 0.5= x but I’m not even sure if that’s right or where to go from there :/
    You're most welcome.

    To rearrange the equation, I would do this:
    (1/√2)/(4-x) = 0.5/x
    1/(√2(4-x)) = 1/(2x)
    2x = √2(4-x) <-- can equate denominators as numerators are both 1
    2x = 4√2 - x√2
    2x + x√2 = 4√2
    x(2+√2) = 4√2
    x = (4√2)/(2+√2)
    rationalise the fraction by multiplying top and bottom by (2-√2) * (2-√2)
    x = (8√2 - 8)/(4-2) = 4√2 -4 = 4(√2 - 1) as required
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    (Original post by FryOfTheMann)
    You're most welcome.

    To rearrange the equation, I would do this:
    (1/√2)/(4-x) = 0.5/x
    1/(√2(4-x)) = 1/2x
    2x = √2(4-x) <-- can equate denominators as numerators are both 1
    2x = 4√2 - x√2
    2x + x√2 = 4√2
    x(2+√2) = 4√2
    x = (4√2)/(2+√2)
    rationalise the fraction by multiplying top and bottom by (2-√2) * (2-√2)
    x = (8√2 - 8)/(4-2) = 4√2 -4 = 4(√2 - 1) as required
    Thanks so much
 
 
 
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