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Teaching plan of math square in the second day of junior high school
This problem can be skillfully solved by rotation method.

Solution: Rotate △PBC 90 degrees counterclockwise around point B until BC and AB overlap, and get a new △AQB, which shows: BQ=PB=2, QA=PC=3, ∠ABQ=∠PBC.

Since ∠ PBC+∠ ABP = 90, ∠ PBQ = ∠ ABQ+∠ ABP = ∠ PBC+∠ ABP = 90, then △PBQ is an isosceles right triangle.

Therefore: ∠ bpq = 45,

Pq2 = Pb2+Bq2 = 2 2+2 2 = 8 is obtained from Pythagorean theorem.

In addition, in △APQ, pa 2+pq 2 =12+8 = 9 = QA 2. According to Pythagorean theorem, △APQ is a right triangle with a right angle of ∠APQ, that is ∠ apq = 90.

To sum up: ∠ APB = ∠ APQ+∠ BPQ = 90+45 =135.