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Homework Help: Finding Triangle Angles with Cosine Rule

  1. Jul 24, 2012 #1
    For part (i), my answer is correct but my answer for (ii) seems to be a little bit out. I can't spot where I've gone wrong. Can anyone help me out?

    Many thanks.

    1. The problem statement, all variables and given/known data

    Q. In the given triangle, find (i) [itex]|\angle abc|[/itex], (ii) [itex]|\angle bac|[/itex].

    3. The attempt at a solution

    (i) cos B = [itex]\frac{c^2+a^2-b^2}{2ac}[/itex] => cos B = [itex]\frac{5^2+8^2-6^2}{(2)(8)(5)}[/itex] => cos B = [itex]\frac{53}{80}[/itex] => B = secant 0.6625 => B = [itex]48^o 31'[/itex]

    (ii) [itex]\frac{sin48^o 31'}{6} = \frac{sin x}{8}[/itex] => [itex]\frac{0.7491}{6} = \frac{sin x}{8}[/itex] => sin x = (8)(0.1249) => x = cosec 0.9989 => x = [itex]87^o42'[/itex]

    Ans.: (From text book): (i) [itex]48^o 31'[/itex], (ii) [itex]92^o 52'[/itex]
     

    Attached Files:

  2. jcsd
  3. Jul 24, 2012 #2
    I'm confused with cos B = 5380 => B = secant 0.6625
    I think you meant arccos? secant is the inverse of cosine, so secant = 1 / cosine
    arccos is the inverse function of cosine, so arccos(cos(x)) = x


    Also, the answer given is larger than 90 degrees. The range of arcsin is -90 to 90, therefore you have to consider all possible results to include other possible angles.
     
  4. Jul 24, 2012 #3
    So arcsin, with range [itex]-90^o[/itex] to [itex]90^o = 180^o[/itex]. Thus, I noticed that [itex]180^o-87^o42'=92^o18'[/itex], which takes me much closer to the intended answer. But am I solving this correctly...?
     
  5. Jul 24, 2012 #4
    I don't have a calculator on me, but that small difference might just be a rounding error. Did you use an exact value when you took arcsin, or the rounded 0.9989?
     
  6. Jul 24, 2012 #5
    Ok, I've gone back to get the exact answer and [itex]\frac{sin48^o31'}{6}=\frac{sinx}{8}[/itex] becomes [itex]x=87^o16'[/itex]. So, [itex]180^0-87^o16'=92^o44'[/itex]. Thats still a little bit out...is that acceptable?
    Also, one other question, what indicator is there to know that I'm meant to subtract [itex]180^0-87^o16'[/itex] and not just use [itex]87^o16'[/itex]
     
  7. Jul 24, 2012 #6

    ehild

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    Homework Helper

    Apply cosine law again to the angle opposite to the side of length 6.

    ehild
     
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