duci BF perpendicular pe DC si CE perpendicular pe AD
a)cosc=0,6=FC/BC, rezulta CF=0,6BC, rezulta ca DC=88=2CF+AB=1,2BC+AB, Notam AB=BC=x, Rezulta 88=2,2x sau
x=88:2,2=40, CF=0,6BC=24
Pt a afla DB folosesti Pitagora generalizata:
DB=radical(BCpatrat+DCpatrat-2BCxDCxcosC)=radical(1600+7744-2x40x88xo,6)=radical9766,4
Ptaafla EC sc rii aria triunghiului ADC in 2 moduri:DCxBF/2=ADxEC/2, sau 88xBF=40xEC
dar BF=radical(BCpatrat-FCpatrat)=radical(1600-576)=radical1024=32
88x32=40xEC, EC=352:5
