ABC, dreptunghic, ∡A=90°, AB=16, ∡B=60°; construim: AD_I_BC. In ΔABC cosB=AB/BC ⇒
BC=AB/cos60°=16:(1/2)=32 cm.=ipotenuza, a) mediana ipotenuzei =1/2ip=(1/2)32= 16 cm
b) In ΔABD dreptunghic in D, avem sinB=AD/AB ⇒ AD=ABsin60°=16*(√3/2)=
AD=8√3 cm=inaltimea corespunzatoare ipotenuzei.