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將△APC繞點C逆時針方向旋轉至△BP'C,連PP',所以∠APC=BP'C,BP'=BP=3顯然△CPP'是等邊三角形,所以∠PP'C=60,PC=PP'=4,在△BPP'中,BP=5,PP'=4,BP'=3,BP^2=PP'^2+PP'^2所以△BPP'是直角三角形,所以∠BP'P= 90°,所以∠B…
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