If the line AP is perpendicular to the plane ABCD through the vertex a of the square ABCD, and AP = AB, then what is the dihedral angle between the plane ABP and the plane CDP? (Figure)

This problem can be calculated by projection method. Let the side length of a square be 1, and the included angle be θ. It means that the projection of ⊿ PCD on plane PAB is ⊿ PAB
Cos θ = s ⊿ PAB ⊿ s ⊿ PCD (⊿ PCD is right triangle, PC, CD is right side)
=（1×1/2）÷（1×√2/2）
=√2/2
Therefore, θ = 45 degree

If the line AP is perpendicular to the plane ABCD through the vertex a of the square ABCD, and AP = AB, then what is the dihedral angle between the plane ABP and the plane CDP? (Figure)

If the line AP is perpendicular to the plane ABCD through the vertex a of the square ABCD, and AP = AB, then what is the dihedral angle between the plane ABP and the plane CDP? (Figure)

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