№7.В треугольнике АВС угол В – прямой. Найдите АС, если: 1) cos A = 0,8, BC=18; 2) sin A = \frac{5}{13}, BA= 36; 3) tgA = 0,75, BA = 8; 4) tgA = 2,4, BC =12.

1) Если cosA = 0.8 = \frac{4}{5}, то \frac{AB}{AC} = \frac{4}{5}. Найдем AB = BC / tgA. \(sinA = \sqrt{1 - cos^2A} = \sqrt{1 - 0.64} = 0.6\) tgA = sinA/cosA = 0.6/0.8 = 3/4. AB = BC / tgA = 18 / (3/4) = 18*4/3 = 24. \(AC = \sqrt{AB^2 + BC^2} = \sqrt{24^2 + 18^2} = \sqrt{576 + 324} = \sqrt{900} = 30\). 2) Если sinA = \frac{5}{13}, то cosA = \sqrt{1 - sin^2A} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}. tgA = \frac{sinA}{cosA} = \frac{5/13}{12/13} = \frac{5}{12}. BC = AB*tgA = 36*5/12 = 15. \(AC = \sqrt{AB^2 + BC^2} = \sqrt{36^2 + 15^2} = \sqrt{1296+225} = \sqrt{1521} = 39\). 3) Если tgA = 0.75 = 3/4, то BC = AB * tgA = 8 * 3/4 = 6. \(AC = \sqrt{AB^2 + BC^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10\). 4) Если tgA = 2.4 = 12/5, то AB = BC / tgA = 12 / (12/5) = 5. \(AC = \sqrt{AB^2 + BC^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\). Ответ: 1) AC = 30. 2) AC = 39. 3) AC = 10. 4) AC = 13.