larissamathiasolivei 16/02/2017 θc = 120° θc / 5 = θf - 32 / 9 120 / 5 = θf - 32/9 5θf - 160 = 1080 5θf = 1080 +160 5θf = 1240 θf= 1240/5 θf= 248°
larissamathiasolivei
θc = 120°
θc / 5 = θf - 32 / 9
120 / 5 = θf - 32/9
5θf - 160 = 1080
5θf = 1080 +160
5θf = 1240
θf= 1240/5
θf= 248°