Тип 1 № 11373

Найдем катеты: $$\tg A = \frac{BC}{AC} = \frac{1}{5} \Rightarrow BC = x, AC = 5x$$ $$AB = \sqrt{AC^2 + BC^2} = \sqrt{(5x)^2 + x^2} = \sqrt{26x^2} = x\sqrt{26} = 13 \Rightarrow x = \frac{13}{\sqrt{26}} = \frac{\sqrt{26}}{2}$$ $$AC = 5x = \frac{5\sqrt{26}}{2}, \quad BC = \frac{\sqrt{26}}{2}$$

Найдем высоту $CH$: $$CH = \frac{AC \cdot BC}{AB} = \frac{\frac{5\sqrt{26}}{2} \cdot \frac{\sqrt{26}}{2}}{13} = \frac{\frac{5 \cdot 26}{4}}{13} = \frac{65}{26} = 2,5$$