Through the vertex a of the square ABCD, make the line PA ⊥ plane ABC, and PA = Pb, then the degree of the sharp dihedral angle formed by plane ABC and plane PCD is? Sorry, wrong number. It's not pa = Pb, it's PA = ab

Through the vertex a of the square ABCD, make the line PA ⊥ plane ABC, and PA = Pb, then the degree of the sharp dihedral angle formed by plane ABC and plane PCD is? Sorry, wrong number. It's not pa = Pb, it's PA = ab

The first question is "be" PC.BD Vertical AC and PA, get BD vertical PC.PC Vertical BD, be, get PC vertical DE.BD =De = √ 2 / √ 3ob / be = √ 2 / 2 / (√ 2 / √ 3) = √ 3 / 2 = sin60 ° angle, bed = 2 * 60 ° = 120 ° second question, make PQ / / CD, PQ = cdpq