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

Решение: а) cosA = 0,6, BA = 12 cosA = BA/AC 0,6 = 12/AC AC = 12 / 0,6 = 20 б) cosA = 0,8, BC = 18 cosA = BA/AC sin^2(A) + cos^2(A) = 1 sinA = sqrt(1 - cos^2(A)) = sqrt(1 - 0,8^2) = sqrt(1 - 0,64) = sqrt(0,36) = 0,6 tgA = sinA / cosA = 0,6 / 0,8 = 3/4 tgA = BC/BA 3/4 = 18/BA BA = 18 * 4/3 = 24 AC = sqrt(BA^2 + BC^2) = sqrt(24^2 + 18^2) = sqrt(576 + 324) = sqrt(900) = 30 в) sinA = 5/13, BC = 10 sinA = BC/AC 5/13 = 10/AC AC = 10 * 13/5 = 26 г) sinA = 5/13, BA = 36 sinA = BC/AC cosA = sqrt(1 - sin^2(A)) = sqrt(1 - (5/13)^2) = sqrt(1 - 25/169) = sqrt(144/169) = 12/13 tgA = sinA / cosA = (5/13) / (12/13) = 5/12 tgA = BC/BA 5/12 = BC/36 BC = 36 * 5/12 = 15 AC = sqrt(BA^2 + BC^2) = sqrt(36^2 + 15^2) = sqrt(1296 + 225) = sqrt(1521) = 39 д) tgA = 0,75, BA = 8 tgA = BC/BA 0,75 = BC/8 BC = 0,75 * 8 = 6 AC = sqrt(BA^2 + BC^2) = sqrt(8^2 + 6^2) = sqrt(64 + 36) = sqrt(100) = 10 е) tgA = 2,4, BC = 12 tgA = BC/BA 2,4 = 12/BA BA = 12 / 2,4 = 5 AC = sqrt(BA^2 + BC^2) = sqrt(5^2 + 12^2) = sqrt(25 + 144) = sqrt(169) = 13