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y= A.4.6 B.13.9 C.21.6
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\[sin(\theta)=\frac{Opp}{Hyp}\] where \(\theta=60\)degrees, \(Opp=y\), and \(Hyp=x\)
you can get one of the sides by using either of \[tan(\theta)=\frac{Opp}{Adj}\] or \[cos(\theta)=\frac{Adj}{Hyp}\] with \(Adj=8\)
I still don't understand how I figure out what y equals so is opp/hyp suppose to =60?
\(y=Opp\)in this triangle, and you can use \(8=Adj\), then you can plus them into the \(tan(\theta)\) equation, in which \(\theta=60\)
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So you add tan and adj?
no...
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