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tan 2B = cot (B - 12)?
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Its either 26 52 or 38
tan(a+b)=tana+tanb/1-tana*tanb
\[\frac{ 1 }{ \tan \theta} =\cot \theta\]
well i have choices in my home work 26 52 38 i cant figure it out i think i have to simplify it
\[\tan \theta=\cot (90-\theta)\]
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26 52 38 are my only options do i have to substitute?
\[ \tan 2B=\tan (90-B+12)\]
cot(B-12)= tan (90-B-12) so 90-B-12 =2B
\[2B=102-B\]
3B=102 B=102/3 = ?
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34
close enough but thanks
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