In rectangular parallelepiped ABCDA1B1C1D1 find the angle between plane A1BC and line BC1, if AA1 = 8, AB = 6, BC = 15.

1. It is necessary to lower the perpendicular С1Н to the straight line СD1.

2. The segment С1Н is perpendicular to any straight line passing through the point Н lying in the given plane (property).

3. Hence <C1HB = 90 ° and the desired angle is the angle С1ВН – the angle between the inclined ВС1 and its projection ВН on the A1BC plane.

4. According to Pythagoras:
D1C = √ (D1C1² + CC1²) = √ (36 + 64) = 10.
BC1 = √ (BC² + CC1²) = √ (225 + 64) = 17.

5. The height С1Н from the right angle is equal to:
C1H = (C1D1 * CC1 / D1C) = 6 * 8/10 = 4.8.

5.sin a:
sin α = C1H / BC1 = 4.8 / 17 ≈ 0.2823.

6.a:
α = arcsin0.2823 ≈ 16.4 °.

Answer: 16.4 °.