[tex]\displaystyle\\
\text{Se da:}\\
\Delta ABC \text{ cu }~m(\sphericalangle A)=90^o ,~ BC=40~\text{cm}~\text{ si }~\sin C= \frac{4}{5}\\\\
\text{Se cere:}\\
a)~AB,~b)~AC,~c)~\sin B,~d)~\cos C,~e)~\text{tg }C,~f)~\text{ctg }C\\\\
\text{Rezolvare:}\\\\
a)\\
\sin C= \frac{4}{5}= \frac{AB}{BC} \\\\
\Longrightarrow~~AB= \frac{4\times BC}{5}= \frac{4\times 40}{5}=4\times 8=\boxed{\bf 32~cm}\\\\
b)\\
AC= \sqrt{BC^2-AB^2}= \sqrt{40^2-32^2}=\\\\
=\sqrt{1600-1024}=\sqrt{576}=\boxed{\bf 24~cm}\\\\
[/tex]
[tex]\displaystyle\\
c)\\
\sin B = \frac{AC}{BC}= \frac{24}{40}= \frac{8\times 3}{8\times 5}=\boxed{\bf \frac{3}{5}}\\\\
d)\\
\cos C= \frac{AC}{BC}= \frac{24}{40}= \frac{8\times 3}{8\times 5}=\boxed{\bf \frac{3}{5}}\\\\
\text{Din c) si d) putem observa ca }~\sin B =\cos C.\\
\text{Explicatia: unghiurile B si C sunt unghiuri complementare.}\\\\
e)\\
\text{tg }C = \frac{AB}{AC}= \frac{32}{24}= \frac{8\times 4}{8\times 3}= \boxed{\bf \frac{4}{3}}\\\\
[/tex]
[tex]\displaystyle\\
f)\\
\text{ctg }C = \frac{AC}{AB}= \frac{24}{32}= \frac{8\times 3}{8\times 4}= \boxed{\bf \frac{3}{4}}\\\\[/tex]
