If the vertex a of the square ABCD passes through the AP vertical plane ABCD and AP = AB, then the degree of the dihedral angle between the plane ABP and the plane DCP is Draw a picture
PA⊥AB PA⊥AD PA=AB=AD
(& nbsp; make BQ vertical plane ABCD & nbsp; & nbsp; BQ = AB & nbsp; & nbsp; & nbsp; link PQ & nbsp; & nbsp; & nbsp; CQ & nbsp; & nbsp; have PQ ⊥ PA & nbsp; PQ ⊥ QB & nbsp; & nbsp; & nbsp; PD ⊥ PQ & nbsp; & nbsp;) & nbsp;
∠ APD is the plane angle of dihedral angle formed by plane ABP and plane DCP