Răspuns :
a. t. lui Pitagora
BC²=AB²+AC², BC²=16+3=25, BC=5
t catetei pentru AC AC²=DCxBC, DC=AC²/BC=9/5
BD=BC-DC=5-9/5=16/5
t inaltimii AD²=BDxDC=16/5x9/5, AD=12/5
b. A=ABxAC/2=4x3/2=6, P=AB+AC+BC==4+3+5=12
c. sinB=AC/BC=3/5, cosB=AB/BC=4/5, tgB=AC/AB=3/4, ctgB=AB/AC=4/3