Тип 1 № 11369

Найдем гипотенузу $AB$: $$\sin A = \sqrt{1 - \cos^2 A} = \sqrt{1 - \left(\frac{\sqrt{35}}{6}\right)^2} = \sqrt{1 - \frac{35}{36}} = \frac{1}{6}$$ $$AB = \frac{BC}{\sin A} = \frac{3}{\frac{1}{6}} = 18$$

Найдем катет $AC$: $$AC = AB \cdot \cos A = 18 \cdot \frac{\sqrt{35}}{6} = 3\sqrt{35}$$

Найдем высоту $CH$: $$CH = \frac{AC \cdot BC}{AB} = \frac{3\sqrt{35} \cdot 3}{18} = \frac{\sqrt{35}}{2}$$

Найдем $AH$: $$AH = \sqrt{AC^2 - CH^2} = \sqrt{(3\sqrt{35})^2 - \left(\frac{\sqrt{35}}{2}\right)^2} =$$ $$= \sqrt{315 - \frac{35}{4}} = \sqrt{\frac{1260 - 35}{4}} = \frac{\sqrt{1225}}{2} = \frac{35}{2} = 17,5$$