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Using the law of cosines, Find c. Triangle ABC, a=8, b=6, y=60
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what is y?
an angle?
is angle y opposite side c?
\(c=a^2+b^2-2ab\cos(C)\) is how this is usually written. my guess is that you have \[c^2=8^2+6^2-2\times 8\times 6\cos(60)\]
the triangle actually does not show where y is
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since \(\cos(60)=\frac{1}{2}\) you get \[c^2=8^2+8^2-8\times 6\] to calculate
sorry i meant \[c^2=8^2+6^2-6\times 8\]
80?
if you are supposed to use the law of cosines, this is the only thing it can be
i get \(c^2=52\) making \(c=\sqrt{52}\)
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52
ok, thanks!
or \(c=2\sqrt{13}\) if you prefer
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