highschoolmom2010
Law of sine question :DD
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AkashdeepDeb
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Why do you have a question if you know what the law of sines is?
highschoolmom2010
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im not sure if i am doing it right
AkashdeepDeb
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You are!
cwrw238
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for a triangle ABC
a/ sinA = b/sinB = c/sinC
highschoolmom2010
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|dw:1377962024865:dw|
\[\frac{ \sin 80 }{ 17 }=\frac{ \sin Z }{ 14 }\]
highschoolmom2010
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like that ????
cwrw238
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yes
highschoolmom2010
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14(sin 80) = 17(sin 17)
highschoolmom2010
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\[\frac{ 14 \sin 80 }{ 17 }=\frac{ 17 \sin Z }{ 17 }\]
highschoolmom2010
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whats where im lost
terenzreignz
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^_^
You're actually doing great :)
Now, 17 cancels out here:
\[\frac{ 14 \sin 80 }{ 17 }=\frac{ \cancel{17} \sin Z }{ \cancel{17} }\]
terenzreignz
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and we get...
\[\Large \sin Z = \frac{14\sin 80^o}{17}\]
highschoolmom2010
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but i plugged the remaining in calculator and got a crazy answer
terenzreignz
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What was it? ^_^
highschoolmom2010
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0.81101814953
highschoolmom2010
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that cant be right
terenzreignz
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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 ^_^
\[\Large Z = \sin^{-1}\left(\frac{14\sin 80^o}{17}\right)\]
highschoolmom2010
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oh i forgot to do teh inverse
terenzreignz
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There's your problem ^_^
I trust it no longer IS a problem? :D
highschoolmom2010
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let me try that and see what i get '
terenzreignz
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I'm sure it'll be more... believable... this time around ^_^
highschoolmom2010
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54.1955269839
terenzreignz
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More reasonable, don't you think? ^_^
terenzreignz
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Correct, by the way :D
highschoolmom2010
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thanks for the clarification :D
terenzreignz
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No problem ^_^