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把△ABP繞點B順時針旋轉60°得到△BCQ,連接PQ,∵∠PBQ=60°,BP=BQ,∴△BPQ是等邊三角形,∴PQ=PB=4,而PC=5,CQ=4,在△PQC中,PQ2+QC2=PC2,∴△PQC是直角三角形,∴∠BQC=60°+90°=150°,∴∠APB=150°.
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