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2014-05-07T07:13:14+03:00
ΔABC, A=90°, AB = 12, cosB = 0.6
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Pabc = ?
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BC =  \frac{AB}{cosB} =  \frac{12}{0.6} = 20
AC =  \sqrt{BC ^{2} -AB ^{2} } = \sqrt{400 -144 } = 16
Pabc = AB + BC + AC = 12 + 20 + 16 = 48