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For this right triangle, what is the cosine of angle C?
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adjacent over hypotenuse
SOH CAH TOA. Sine, Opposite over Hypotenuse Cosine, Adjacent over Hypotenuse Tangent, Opposite over Adjacent
but you don't know the adjacent side yet, you need to find it pythagoras tells you that \[5^2+b^2=13^2\] \[25+b^2=169\] \[b^2=144\] \[b=12\]
or you can just memorize the \(5-12-13\) right triangle, it is a favourite of math teachers
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A: 12/13 B: 13/12 C: 5/12 D: 5/13
b^2 = 13^2 - 5^2 b = sqrt( 169-25) = 12 cos(c) = b/13
only one most of them know besides \(3-4-5\) in any case it is \[\cos(C)=\frac{opp}{hyp}=\frac{12}{13}\]
Wow; that helped a ton. Thank you bunchessss!
I'm horrible in mathh.
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