[tex]\displaystyle \\
\texttt{Se da: } \\
\Delta ABC ~~cu~~\ \textless \ A=90^o \\
BC = 80 cm \\ \\
\frac{AB}{AC} = \frac{4}{3} \\ \\
AD \perp BC ~~~D \in BC~~~(inaltime)\\
\texttt{Se cere: } \\
AB=? \\
AC=? \\
AD=? \\ \\
Rezolvare: \\ \\
\frac{AB}{AC} = \frac{4}{3} \\ \\
\Longrightarrow~~AB =4k~~si~~ AC=3k \\
\texttt{Aplicam Pitagora:} \\
AB^2+AC^2 = BC^2 \\
(4k)^2 + (3k)^2=80^2 \\
16k^2 + 9K^2 = 6400 \\
k^2(16+9) = 6400 \\
25 k^2 = 1600 \\ \\
k^2 = \frac{6400}{25}[/tex]
[tex]\displaystyle \\
k = \sqrt{\frac{6400}{25}} =\frac{80}{5} = \boxed{16} \\ \\
AB = 4k = 4\times 16 = \boxed{64 ~cm} \\
;AC = 3k = 3\times 16 = \boxed{48 ~cm} \\ \\
\texttt{Calculam inaltimea:} \\
BC \times AD =AB \times AC\\ \\
AD = \frac{AB \times AC}{BC}= \frac{64 \times 48}{80}= \frac{8 \times 48}{10}=\frac{384}{10}= \boxed{38,4 ~cm}
[/tex]
