AD _|_BC
din tr ADC ⇒cu Pitagora
AD²=AC²=900-324=576
AD=24 cm =h
A=AD.CB/2=24.36/2=432cm²
BE _|_ AC
A=BE.AC/2=432
30.BE=432.2
BE=28,8 cm
b) cos BAC daca notam BAD=x⇒ BAC=2x
cos 2x=2cosx.sinx
cosx=24/30
sinx 18/30
cos BAC=2.24/30.18/30=216/225
c)r=A/p
A-aria
p-semiperimetrul
A=432
p=96/2=48
r=432/48=9 cm