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由正弦定理得:BC/sinA=1/sin45°又∵角A=180°-45°-角B∴BC/sin(135°-角B)=根號2∴BC=根號2·sin(135°-角B)展開得BC=根號2·(根號2/2 cosB+根號2/2 sinB)即BC=根號2·sin(45°+B)∵在此三角形中,0°<…
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