Прямоугольные треугольники

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

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

1. Найдем катет $AC$: $$AC = AB \cdot \cos A = 27 \cdot \frac{2}{3} = 18$$

2. Найдем AH: $$AH = AC\cos A = 18\cdot\frac{2}{3} = 12$$

$$BH = AB - AH$$

$$AH = AC\cos A$$

1. Найдем катет $AC$: $$AC = AB \cdot \cos A = 27 \cdot \frac{2}{3} = 18$$

2. Найдем AH: $$AH = AC\cos A = 18\cdot\frac{2}{3} = 12$$

3. Находим BH:

$$BH = 27 - 12 = 15$$

1. Найдем катет $AC$: $$AC = AB \cdot \cos A = 4 \sqrt{15} \cdot 0,25 = \sqrt{15}$$

2. Найдем катет $BC$: $$BC = \sqrt{AB^2 - AC^2} = \sqrt{(4 \sqrt{15})^2 - (\sqrt{15})^2} = \sqrt{240 - 15} = \sqrt{225} = 15$$

3. Найдем высоту $CH$:

$$Sabc = \frac{1}{2}AC\cdot CB = \frac{1}{2} CH \cdot AB \Leftrightarrow CH = \frac{AC\cdot CB}{AB}$$

$$CH = \frac{AC \cdot CB}{AB} = \frac{\sqrt{15} \cdot 15}{4 \sqrt{15}} = \frac{15}{4} = 3,75$$

$$AH = AC\cos A$$

$$AC = AB\cos A \Rightarrow AH = AB\cos^2 A$$

Воспользуемся формулой: $\cos^2a = \frac{1}{1 + \tg^2a}$

$$AH = 13\cdot\frac{1}{1 + \left(\frac{1}{5}\right)^2} = 13\cdot\frac{25}{26} = 12,5$$