1.试题呈现题目如图1,AB是圆O的直径,PA垂直于圆O所在的平面,C是圆周上不同于A、B的任意一点.设AB=AP=2,AC=1,且AD⊥PC,垂足为点D,求二面角D-AB一P的大小.分析:已知PA⊥平面ABC,可得PA⊥BC.又AC⊥BC,则BC⊥平面PAC,从而AD⊥BC.又AD⊥PC,所以AD⊥平面PBC.因此,本题至少包含三个线面垂直的结构,通过线面垂直,可以得到三棱锥P-ABC的各面以及△ADC均是直角三角形,给计算提供了很大方便. 1. Questions presented in Figure 1, AB is the diameter of the circle O, PA is perpendicular to the plane where the circle O, C is any point on the circumference different from A, B. Let AB = AP = 2, AC = 1, and AD ⊥ PC, vertical foot for the point D, seeking dihedral angle D-AB-P size analysis: known PA ⊥ plane ABC, available PA ⊥ BC and AC ⊥ BC, BC ⊥ plane PAC, which AD Therefore, the problem contains at least three vertical lines of the structure, through the vertical line, you can get the surface of the triangular pyramid P-ABC and △ ADC are right-angled triangle, To the calculation provided a great convenience.