daca AD=5 atunci BC = 2AD =15
Δ ADC echilateral mas<C=60
BA² = BC² - AC²
AB² = 10² - 5² = 100 - 25 =75
AB =√25√3 = 5√3
peri.= BC+AC+AB= 10 +5+5√3 = 5(3+√3)
aria = AC·AB/2 =5·5√3/2 = 25√3/2
sinC= AB / BC= 5√3/10 =√3/2 ⇒C=60
cosC= AC/BC = 5/10 = 1/2
tgC= AB/AC= 5√3 /5 = √3