Finding Triangle Angles with Cosine Rule

  • Thread starter odolwa99
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  • #1
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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.

Homework Statement



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

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]
 

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Answers and Replies

  • #2
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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.
 
  • #3
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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...?
 
  • #4
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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?
 
  • #5
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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]
 
  • #6
ehild
Homework Helper
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Apply cosine law again to the angle opposite to the side of length 6.

ehild
 

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