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Solve the triangle: A = 51°, b = 14, c = 6
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@Loser66
@e.mccormick i just need a, B, and C
|dw:1371668687548:dw|
or else?
law of cosines gives us \[a^2=b^2+c^2-2bc~cosA\] the rest is elementary
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Yeah thats it, I just need a C and B
232 - 2(14)(6)cos51 = a^2 How do i do it?
solving for the angle part: \[C=cos^{-1}\left(\frac{c^2-a^2-b^2}{-2ab}\right)\] \[B=cos^{-1}\left(\frac{b^2-a^2-c^2}{-2ac}\right)\]
you generally do it with a calculator
but how do i do this? 232 - 2(14)(6)cos51 = a^2
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http://www.wolframalpha.com/input/?i=sqrt%2814%5E2%2B6%5E2-2%286%29%2814%29*cos%2851%29%29
Thanks! Just that one value gave me the other ones.
:)
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