在长方体ABCD-A'B'C'D'中,AB=4,BC=AA'=3,直线BD'与平面BB'C'C所成的角的大小

在长方体ABCD-A'B'C'D'中,AB=4,BC=AA'=3,分别求直线BD'与平面ABCD,直线BD'与平面BB'C'C所成的角的大小

---

答蠢誉案是arctan[(2/3)*√2]

根据长方体的定义及公理:线DC'与面BB'C'C垂直,
那么直角三角形BC'D'中∠D'BC'就是所求的角.
根据已知条件,三角带困段形BC'D'中,线BC'=3*√2,线C'尺高D'=4,
所以∠D'BC'=arctan[(2/3)*√2]

---

1.直线BD'简空与平面ABCD所成角=arctan(DD'/BD)
=arctan[3/√(3²+4²)]
=arctan3/5

2.直线BD'与平面BB'C'桥咐顷C所成的角敏陆=arctan(C'D'/BC')
=arctan[4/√(3²+3²)]
=arctan(2√2/3)

---