1), after running Xs, PC=4-x, CQ=2x. Because PQ is perpendicular to AC, triangle PCQ is a right triangle with angle C=60 degrees and angle QPC=30 degrees. Therefore, PC=2CQ is obtained, that is, 4-x=2*2x and x = 0.8s..

2) When 0

3) Let PQ and AD intersect at point E, the vertical line passing through point Q is AD, and intersect with AD at point F, and the angle DAC=30 degrees, then QF= 1/2AQ = 4-2x, then QF=2-x=PD, and the area of triangular PDE = 1/2*DE*PD.