m∡(BOE)=60°=2m∡(CAB) ∡(BOE, unghi exterior,AOBtr.isoscel)⇒m∡9CAB)=30°⇒
⇒(teorema unghiulide 30°)BC=AC/2=24/2=12dam
AB= (Pitagora tr dr ABC)=√(24²-12²)=12√3dam, cerinta
b) distanta maxima=FC= (Pitagora in tr.dr FBC)=√(FB²+BC²)=√((6√3)²+12²)=
=6√(3+4)=6√7dam=60√7m≈158,74m≈159m, cerinta
am folosit FB=AB/2=12√3:2=6√3 dam
c) ΔFEB≈ΔCED unghi unghi,∡FEB≡∡CED, opuse la varf,∡EFB≡∡ECD, alterne interne
FB/DC=1/2=EB/DE (1)
cum EB+DE=DB=AC=24 si tinand cont de (1)⇒3EB=DB⇒EB=DB/3=8dam, cerinta
