in ΔABC :
sin B = cateta opusa/ipotenuza
3/4 = AC/BC
3/4 = AC/10 => AC = (3×10):4 = 7,5 cm
Aplicam Teorema lui Pitagora :
AB^2 + AC^2 = BC^2 => AB^2 =BC^2 - AC^2
Din formula a^2-b^2=(a-b)(a+b) => AB^2 = (BC-AC)(BC+AC)
=> AB^2 = 2,5 × 17,5 = 43,75 => AB = \/43,75 ~ 6,62 cm.
P abc = AB +AC + BC = 6,62 + 7,5 + 10 = 24,12cm.
A dat ciudat, dar m am verificat si cred ca e corect :*
