[tex]\it \Delta ABC-dr, \hat A=90^o,\ \stackrel{T.P.}{\Longrightarrow}\ BC^2=AB^2+AC^2=15^2+20^2=\\ \\ =225+400=625=25^2 \Rightarrow BC=25\ cm[/tex]
[tex]\it sinB=\dfrac{cateta\ \ opus\breve a}{ipotenuz\breve a}=\dfrac{AC}{BC}=\dfrac{\ 20^{(5}}{25}=\dfrac{4}{5}\\ \\ \\ cosB=\dfrac{cateta\ \ alaturat\breve a}{ipotenuz\breve a}=\dfrac{AB}{BC}=\dfrac{\ \ 15^{(5}}{25}=\dfrac{3}{5}[/tex]
[tex]\it tgC=\dfrac{cateta\ \ opus\breve a}{cateta\ alaturat\breve a}=\dfrac{AB}{AC}=\dfrac{\ \ 15^{(5}}{20}=\dfrac{3}{4}\\ \\ \\ ctgC=\dfrac{1}{tgC}=\dfrac{4}{3}[/tex]
