[tex]Zadanie\\\\c=b+4\\sin\alpha=\frac{5}{13}\\cos\alpha=?\\\\sin^{2}\alpha+cos^{2} \alpha=1\\\\(\frac{5}{13})^{2}+cos^{2} \alpha=1\\\\\frac{25}{169}+cos^{2} \alpha=1\\\\ cos^{2} \alpha=1-\frac{25}{169}\\\\cos^{2} \alpha=\frac{144}{169}\\\\cos\alpha= \sqrt{\frac{144}{169} }\\\\ cos\alpha= \frac{12}{13}\\\\\frac{b}{b+4}=cos\alpha \\\\\frac{b}{b+4}=\frac{12}{13}\\\\13b= 12(b+4) \\\\13b=12b+48\ \ \mid-12b\\\\b= 48\\\\c=b+4\\\\c=48+4=52\\\\a=?[/tex]
[tex]\\\\a^{2}+b^{2}= c^{2}\\\\a^{2}+48^{2}=52^{2}\\\\a^{2}+2304=2704\ \ \mid-2304\\\\ a^{2} =400\\\\ a=\sqrt{400}\\\\a=20\\\\Ob._{\vartriangle}=a+b+c\\\\ Ob._{\vartriangle} = 20+48+52=120[/tex]