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Mathematics 19 Online
OpenStudy (anonymous):

For this right triangle, what is the cosine of angle C?

OpenStudy (anonymous):

adjacent over hypotenuse

OpenStudy (anonymous):

OpenStudy (anonymous):

SOH CAH TOA. Sine, Opposite over Hypotenuse Cosine, Adjacent over Hypotenuse Tangent, Opposite over Adjacent

OpenStudy (anonymous):

but you don't know the adjacent side yet, you need to find it pythagoras tells you that \[5^2+b^2=13^2\] \[25+b^2=169\] \[b^2=144\] \[b=12\]

OpenStudy (anonymous):

or you can just memorize the \(5-12-13\) right triangle, it is a favourite of math teachers

OpenStudy (anonymous):

A: 12/13 B: 13/12 C: 5/12 D: 5/13

OpenStudy (jhannybean):

b^2 = 13^2 - 5^2 b = sqrt( 169-25) = 12 cos(c) = b/13

OpenStudy (anonymous):

only one most of them know besides \(3-4-5\) in any case it is \[\cos(C)=\frac{opp}{hyp}=\frac{12}{13}\]

OpenStudy (anonymous):

Wow; that helped a ton. Thank you bunchessss!

OpenStudy (anonymous):

I'm horrible in mathh.

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