9 В треугольнике АВС угол C равен 90°, СН — высота, AB = 45, sin A = 1/3. Найдите длину отрезка АН.

Объяснение:

1. В прямоугольном треугольнике ABC:
• Given: \( AB = 45 \), \( \sin A = \frac{1}{3} \)
• We know that \( \sin A = \frac{\text{противолежащий катет}}{\text{гипотенуза}} \). In \( \triangle ABC \), the side opposite to angle A is BC, and the hypotenuse is AB.
• So, \( \sin A = \frac{BC}{AB} \)
• \( \frac{1}{3} = \frac{BC}{45} \)
• \( BC = 45 \times \frac{1}{3} = 15 \)
2. Now consider the right-angled triangle ACH. Angle ACH is 90 degrees, and CH is the altitude.
3. In \( \triangle ABC \), we can find \( \cos A \) using the identity \( \sin^2 A + \cos^2 A = 1 \).
4. \( \cos^2 A = 1 - \left( \frac{1}{3} \right)^2 = 1 - \frac{1}{9} = \frac{8}{9} \)
5. \( \cos A = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3} \) (Since A is an acute angle in a right-angled triangle, \( \cos A \) is positive).
6. In right-angled triangle ACH, angle A is the same as in triangle ABC.
7. We have \( \cos A = \frac{\text{прилежащий катет}}{\text{гипотенуза}} \). In \( \triangle ACH \), the adjacent side to angle A is AH, and the hypotenuse is AC.
8. So, \( \cos A = \frac{AH}{AC} \).
9. We need to find AC. We can use the Pythagorean theorem in \( \triangle ABC \): \( AC^2 + BC^2 = AB^2 \)
10. \( AC^2 + 15^2 = 45^2 \)
11. \( AC^2 + 225 = 2025 \)
12. \( AC^2 = 2025 - 225 = 1800 \)
13. \( AC = \sqrt{1800} = \sqrt{900 \times 2} = 30\sqrt{2} \)
14. Now we can find AH:
15. \( \frac{AH}{AC} = \cos A \)
16. \( AH = AC \times \cos A \)
17. \( AH = 30\sqrt{2} \times \frac{2\sqrt{2}}{3} \)
18. \( AH = \frac{30 \times 2 \times 2}{3} = \frac{120}{3} = 40 \)

Ответ: 40