sinB=AD/AB=12/AB=3/5⇒AB=12×5/3=20
cu Pitagora BD²=AB²-AD²=400-144=256⇒BD=16
In triunghiul ADC⇒AD²+DC²=AC²
stim ca cosC=DC/AC=5/13
DC=5AC/13
144+25AC²/169=AC²
aducand la acelasi numitor⇒
144×169=169AC²-25AC²
144AC²=144×169
AC²=169
AC=13
DC²=AC²-AD²
DC²=169-144=25
DC=5
BC=DC+BD=5+16=21