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. In Triangle ABC, Cos B = (9/12) .....find Sin B.
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\[B = \cos^{-1}(\frac{9}{12})\]
whered the -1 come from?
means "inverse"
ohhh ok
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\[\text{Sin}[\text{ArcCos}[9/12]]=\frac{\sqrt{7}}{4} \]
wth robtobey
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