sin B = [tex]sin B = \frac{3}{5} = \frac{cateta-opusa}{ipotenuza} = \frac{AC}{BC} \\=> \frac{AC}{BC} = \frac{3}{5}\\\frac{AC}{15} =\frac{3}{5} \\AC = 15 x 3 : 5\\AC = 3 x 3 = 9 cm\\\\Aplicam Teorema-lui-Pitagora:\\BC^{2} = AC^{2} + AB^{2} \\15^{2} = 9^{2} + AB^{2} \\225 = 81 + AB^{2} \\=> AB^{2} = 144 | scoti radical \\=> AB = 12 cm[/tex]
