ΔABC, A=90°, AB = 12, cosB = 0.6
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Pabc = ?
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BC = [tex] \frac{AB}{cosB} [/tex] = [tex] \frac{12}{0.6} [/tex] = 20
AC = [tex] \sqrt{BC ^{2} -AB ^{2} } [/tex] =[tex] \sqrt{400 -144 } [/tex] = 16
Pabc = AB + BC + AC = 12 + 20 + 16 = 48
