∴AP⊥BC and ∵AB⊥BC

∴BC⊥ Aircraft Company

∴BC⊥AD

In RT△PAB, pa = ab, PD = db, then AD⊥PB.

So there is an AD⊥ flat PCB and an AD⊥PC.

Make DE⊥PC and AD⊥PC in the previous step.

Then PC⊥ airplane Pabu has PC⊥AE.

Because de⊥pc∠aed is the size of dihedral angle A-PC-B.

Let's ask ∠AED.

Let PA=AB=BC=6, then AC=6√2 (2 under the root sign) and PC=6√3. AE=2√6，AD=3√2

In RT△ADE, AD=3√2, AE=2√6,

Then sin∠AED=AD/AE=√3 /2, then ∠AED=60?

That is, dihedral angle A-PC-B is 60? .