Задача 80613 решить задание по геометрии ...
Условие
Решение
AD = BC = 5, AA2 = BB2 = 3, B1C2 = A1D2 = 1, A2D2 = B2C2 = 3,
A2B2 = AB = A1B1 = CD = C1D1 = C2D2 = 1
Найти: tg BCB1, cos AB2C, cos AD1C
Решение.
1) B1C1 = BC - B2C2 = 5 - 3 = 2
CC1 = DD1 = BB2 - B1C2 = 3 - 1 = 2
Треугольник B1CC1 - прямоугольный и равнобедренный.
Угол C1CB1 = 45°, Угол BCB1 = BCC1 - C1CB1 = 90° - 45° = 45°
[b]tg BCB1 = tg 45° = 1[/b]
2) По теореме Пифагора:
AB2 = sqrt(AB^2 + BB2^2) = sqrt(1^2 + 3^2) = sqrt(1 + 9) = sqrt(10)
AC = sqrt(AB^2 + BC^2) = sqrt(1^2 + 5^2) = sqrt(1 + 25) = sqrt(26)
B2C = sqrt(BB2^2 + BC^2) = sqrt(3^2 + 5^2) = sqrt(9 + 25) = sqrt(34)
По теореме косинусов:
AC^2 = AB2^2 + B2C^2 - 2*AB2*B2C*cos AB2C
26 = 10 + 34 - 2*sqrt(10)*sqrt(34)*cos AB2C
2*sqrt(340)*cos AB2C = 44 - 26 = 18
cos AB2C = 18/(2sqrt(340)) = 9/sqrt(340)
[b]cos AB2C = 9sqrt(340)/340[/b]
3) По теореме Пифагора:
AC = sqrt(AB^2 + BC^2) = sqrt(1^2 + 5^2) = sqrt(1 + 25) = sqrt(26)
AD1 = sqrt(AD^2 + DD1^2) = sqrt(5^2 + 2^2) = sqrt(25 + 4) = sqrt(29)
CD1 = sqrt(CD^2 + DD1^2) = sqrt(1^2 + 2^2) = sqrt(1 + 4) = sqrt(5)
По теореме косинусов:
AC^2 = AD1^2 + CD1^2 - 2*AD1*CD1*cos AD1C
26 = 29 + 5 - 2*sqrt(29)*sqrt(5)*cos AD1C
2*sqrt(145)*cos AD1C = 34 - 26 = 8
cos AD1C = 8/(2*sqrt(145)) = 4/sqrt(145)
[b]cos AD1C = 4sqrt(145)/145[/b]
2 Задача.
AB = CD = 6, A1A2 = D1D2 = 2, B1B2 = C1C2 = 2, A3B3 = C3D3 = 2
AA1 = BB1 = CC1 = DD1 = 2, A2A3 = B2B3 = C2C3 = D2D3 = 2
AD = A1D1 = A2D2 = A3D3 = BC = B1C1 = B2C2 = B3C3 = 1
Найти: tg BAA3, cos AB1C, cos AD1A2
Решение.
1) Смотрите рисунок. Проводим высоту A2H, обозначена красным.
AH = A1A2 = 2, A2H = AA1 = 2, A3H = A2H + A2A3 = 2 + 2 = 4
tg BAA3 = tg HAA3 = A3H/AH = 4/2
[b]tg BAA3 = 2[/b]
2) По теореме Пифагора:
AC = sqrt(AB^2 + BC^2) = sqrt(6^2 + 1^2) = sqrt(36 + 1) = sqrt(37)
AB1 = sqrt(AB^2 + BB1^2) = sqrt(6^2 + 2^2) = sqrt(36 + 4) = sqrt(40)
B1C = sqrt(BB1^2 + BC^2) = sqrt(2^2 + 1^2) = sqrt(4 + 1) = sqrt(5)
По теореме косинусов:
AC^2 = AB1^2 + B1C^2 - 2*AB1*B1C*cos AB1C
37 = 40 + 5 - 2*sqrt(40)*sqrt(5)*cos AB1C
2*sqrt(200)*cos AB1C = 45 - 37 = 8
cos AB1C = 8/(2*sqrt(200)) = 4/(10*sqrt(2))
[b]cos AB1C = 2sqrt(2)/5[/b]
3) По теореме Пифагора:
AA2 = sqrt(AA1^2 + A1A2^2) = sqrt(2^2 + 2^2) = sqrt(4 + 4) = sqrt(8)
AD1 = sqrt(AD^2 + DD1^2) = sqrt(1^2 + 2^2) = sqrt(1 + 4) = sqrt(5)
D1A2 = sqrt(D1A1^2 + A1A2^2) = sqrt(2^2 + 2^2) = sqrt(4 + 4) = sqrt(8)
По теореме косинусов:
AA2^2 = AD1^2 + D1A2^2 - 2*AD1*D1A2*cos AD1A2
8 = 5 + 8 - 2*sqrt(5)*sqrt(8)*cos AD1A2
2*sqrt(40)*cos AD1A2 = 13 - 8 = 5
cos AD1A2 = 5/(2sqrt(40)) = 5/(4sqrt(10)) = 5sqrt(10)/(4*10)
[b]cos AD1A2 = sqrt(10)/8[/b]
