Fie <A=90, AB>AC si M apartine lui AC astfel incat AM=4cm si BM=5cm.
Aplicam teorema bisectoarei:
cos(<B)=AB/BC=AM/BM=4/5
sin(<B)=AC/BC=9/BC
(sinH)^2 +(cosH)^2=1
(sinB)^2 +(cosB)^2=1
(4/5)^2 +(9/BC)^2 =1
16/25 +81/BC^2 =1
81/BC^2=9/25
9/BC=3/5
BC=15cm, stim ca AC=9cm, aplicam T. lui Pitagora:
AB^2=BC^2 -AC^2=225-81=144
AB=12cm
