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Why do you have a question if you know what the law of sines is?
im not sure if i am doing it right

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Other answers:

You are!
for a triangle ABC a/ sinA = b/sinB = c/sinC
|dw:1377962024865:dw| \[\frac{ \sin 80 }{ 17 }=\frac{ \sin Z }{ 14 }\]
like that ????
yes
14(sin 80) = 17(sin 17)
\[\frac{ 14 \sin 80 }{ 17 }=\frac{ 17 \sin Z }{ 17 }\]
whats where im lost
^_^ You're actually doing great :) Now, 17 cancels out here: \[\frac{ 14 \sin 80 }{ 17 }=\frac{ \cancel{17} \sin Z }{ \cancel{17} }\]
and we get... \[\Large \sin Z = \frac{14\sin 80^o}{17}\]
but i plugged the remaining in calculator and got a crazy answer
What was it? ^_^
0.81101814953
that cant be right
Ahh... I see. Because what you got was the SINE of the angle Z :) You have to run it through \(\large \sin^{-1}\) to get the angle Z itself ^_^ \[\Large Z = \sin^{-1}\left(\frac{14\sin 80^o}{17}\right)\]
oh i forgot to do teh inverse
There's your problem ^_^ I trust it no longer IS a problem? :D
let me try that and see what i get '
I'm sure it'll be more... believable... this time around ^_^
54.1955269839
More reasonable, don't you think? ^_^
Correct, by the way :D
thanks for the clarification :D
No problem ^_^

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