Fie {D} proiectia lui A pe BC⇒ AD- h
⇒AD⊥BC⇒ m(∡ADB)=90°⇒ΔADB-dr
ΔABD-dr : cos ∡ABD=cat al\ip
cos 60°=BD\AB
1\2=BD\10⇒BD=5
*In acelasi triunghi*
sin ∡B=cat op\ip
sin 60°=AD\AB
√3\2=AD\10⇒AD=5√3
ΔADC-dr[tex]T.P.\\===>[/tex]
[tex]DC^{2} =AC^{2} -AD^{2} \\DC^{2} =(10\sqrt{3} )^{2} - (5\sqrt{3} )^{2} \\DC^{2} =300-75\\DC=\sqrt{225} =>DC=15[/tex]
BC=BD+DC
BC=5+15
BC=20
*Acum,ar fi indicat sa verificam daca triunghiul e dreptunghic,deci asta si facem*
*Vom folosi RTP (reciproca teoremei lui Pitagora)
ΔABC:
[tex]BC^{2} =AC^{2} +AB^{2} \\20^{2} =(10\sqrt{3} )^{2} +10^{2} \\400=300+100[/tex]
400=400 (A) [tex]RTP\\==>[/tex] ΔABC-dr
