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Math Help - Sine Rule Help please

  1. #1
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    Sine Rule Help please

    Hi maths gurus.

    I'm in need of some help here with the sine rule.

    I have my sine rule values 3.1/sin20 = 5/sineB = 6/SineC

    I am trying to find the value for the sine of B so what I did was 3.1/Sine(20)*5 to get 1.81 If I do the inverse sine it comes up with an error on my calculator so is this what I'm meant to do or what?
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  2. #2
    Junior Member Bradyns's Avatar
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    Re: Sine Rule Help please

    You want  sin(B) = \frac{5sin(20)}{3.1}

    \frac{3.1}{5sin(20)} will never work, as it is greater than 1.
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  3. #3
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    Re: Sine Rule Help please

    Ah Okay thank you very much! It worked
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  4. #4
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    Re: Sine Rule Help please

    A little more clarification: you started from \frac{3.1}{sin(20)}= \frac{5}{sin(B)} and divided both sides by 5 to get
    \frac{3.1}{5 sin(20)}= \frac{1}{sin(B)}. The right side is \frac{1}{sin(B)}, NOT sin(B) so you cannot immediately take the arcsine. To solve for B, you have to take the reciprocal of both sides to get sin(B)= \frac{5 sin(20)}{3.1}. then take the arcsine.
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