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What is the length of AD? Please Help!!!!!!
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BAM = x so;BAC = 2x since AMis amedian BM =MC and AC = 12 (guven) for AMCtriangel sin2x/MC = sinAMC/12 sinAMC = 12sin2x/MC--------1 for ABM trinagel Sin(180-AMC)/AB = sinx/MB sin(180-AMC) = sinAMC = ABsinx/MB--------2 1 = 2 ABsinx/MB = 12sin2x/MC ABsinx= 12.2.sinx.cosx (b'coz MC=MB] AB/24 = cosx-----3 to ABDtriangel cosx = AB/AD------4 3=4 AD = 24
|dw:1373043992723:dw|
........24
Wrong
o
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well i suck at math
first find ac .... then u will get ad
lol
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