[tex]\it \widehat C+\widehat B=90^o \Rightarrow sinC=cosB=\dfrac{3}{5}\\ \\ cosC=\sqrt{1-sin^2C} =\sqrt{1-\Big(\dfrac{3}{5}\Big)^2}=\sqrt{1-\dfrac{9}{25}}=\sqrt{\dfrac{16}{25}}=\dfrac{4}{5}\\ \\ \widehat{BAD}=\widehat C\ (au\ acela\c{s}i\ complement,\ unghiul\ B) \Rightarrow tg(BAD)=tgC=\dfrac{sinC}{cosC}=\\ \\ \\ =\dfrac{\dfrac{3}{\not5}}{\dfrac{4}{\not5}}=\dfrac{3}{4}[/tex]
[tex]\it \Delta ADC \Rightarrow cosC=\dfrac{CD}{AC} \Rightarrow \dfrac{^{4)}4}{\ 5}=\dfrac{16}{AC} \Rightarrow \dfrac{16}{20}=\dfrac{16}{AC} \Rightarrow AC=20\ cm\\ \\ \\ \Delta ABC \Rightarrow cosC=\dfrac{AC}{BC} \Rightarrow \dfrac{^{5)}4}{\ 5}=\dfrac{20}{BC} \Rightarrow \dfrac{20}{25}=\dfrac{20}{BC} \Rightarrow BC=25\ cm[/tex]
[tex]\it \Delta ABC \Rightarrow sinC=\dfrac{AB}{BC} \Rightarrow \dfrac{^{5)}3}{\ 5}=\dfrac{AB}{25} \Rightarrow \dfrac{15}{25}=\dfrac{AB}{25} \Rightarrow AB=15\ cm[/tex]
[tex]\it \mathcal{P}_{ABC}=AB+AC+BC=15+20+25=60\ cm[/tex]
