[tex]Obw = 60\\oraz\\Obw = a + b + c\\\\ctg\alpha = \frac{4}{3}\\oraz\\ctg\alpha = \frac{a}{b} = \frac{3}{4} \ \ \rightarrow \ \ a = \frac{4}{3}a\\\\\\a^{2}+b^{2} = c^{2}\\\\a^{2}+(\frac{4}{3}a)^{2} = c^{2}\\\\a^{2}+\frac{16}{9}a^{2} = c^{2}\\\\\frac{25}{9}a^{2} = c^{2}\\\\c = \sqrt{\frac{25}{9}a^{2}}\\\\\underline{c = \frac{5}{3}a}[/tex]
[tex]a+b+c = 60\\\\a + \frac{4}{3}a + \frac{5}{3}a = 60\\\\\frac{12}{3}a = 60\\\\4a = 60 \ \ /:4\\\\\underline{a = 15}\\\\b = \frac{4}{3}a = \frac{4}{3}\cdot15\\\\\underline{b = 20}\\\\c = \frac{5}{3}a = \frac{4}{3}\cdot15\\\\\underline{c = 25}[/tex]
Odp. Boki tego trójkąta mają długości: 15, 20, 25.
