[tex] \displaystyle\\
\text{Se da:}\\\\
\Delta ABC~ \text{ dreptunghic cu }~ m(\sphericalangle A)=90^o\\\\
\sin B =\frac{3}{5}\\\\
BC= 15~cm\\\\
\text{Se cere:}\\
\text{Lungimea catetelor AB si AC}\\\\
\text{Rezolvare:}\\\\
\text{Cateta AC este opusa unghiului B.}\\\\
\Longrightarrow~AC=BC\sin B=15\cdot\frac{3}{5}=\frac{15\cdot 3}{5}= \frac{45}{5}=\boxed{\bf 9~cm}\\\\
\text{Cateta AB o calculam cu teorema lui Pitagora.}\\\\
AB = \sqrt{BC^2-AC^2}=\sqrt{15^2-9^2}=\sqrt{225-81}=\sqrt{144}= \boxed{\bf 12~cm} [/tex]