348. В треугольнике ABC угол B прямой. Найдите AC, если: a) cosA=0,6, BA=12; б) cosA=0,8, BC=18; в) sinA = \frac{5}{13}, BC=10; г) sinA = \frac{5}{13}, BA=36; д) tgA=0,75, BA=8; e) tgA=2,4, BC=12.

В прямоугольном треугольнике ABC с прямым углом B, требуется найти гипотенузу AC. а) cosA = 0,6, BA = 12 [cosA = \frac{BA}{AC} \Rightarrow AC = \frac{BA}{cosA} = \frac{12}{0,6} = 20] б) cosA = 0,8, BC = 18 [cosA = \frac{AB}{AC}\] [sinA = \sqrt{1 - cos^2A} = \sqrt{1 - 0.8^2} = \sqrt{1 - 0.64} = \sqrt{0.36} = 0.6] [sinA = \frac{BC}{AC} \Rightarrow AC = \frac{BC}{sinA} = \frac{18}{0.6} = 30] в) sinA = 5/13, BC = 10 [sinA = \frac{BC}{AC} \Rightarrow AC = \frac{BC}{sinA} = \frac{10}{\frac{5}{13}} = \frac{10 \cdot 13}{5} = 26] г) sinA = 5/13, BA = 36 [cosA = \sqrt{1 - sin^2A} = \sqrt{1 - (\frac{5}{13})^2} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}] [cosA = \frac{BA}{AC} \Rightarrow AC = \frac{BA}{cosA} = \frac{36}{\frac{12}{13}} = \frac{36 \cdot 13}{12} = 39] д) tgA = 0,75, BA = 8 [tgA = \frac{BC}{BA} \Rightarrow BC = tgA \cdot BA = 0.75 \cdot 8 = 6] [AC = \sqrt{AB^2 + BC^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10] e) tgA = 2,4, BC = 12 [tgA = \frac{BC}{BA} \Rightarrow BA = \frac{BC}{tgA} = \frac{12}{2.4} = 5] [AC = \sqrt{AB^2 + BC^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13] **Ответы:** a) AC = 20 б) AC = 30 в) AC = 26 г) AC = 39 д) AC = 10 e) AC = 13