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連接AB,根據畢氏定理可以得到OA=OB=10,AB=8根據余弦定理可以得到:OA2+OB2-2OA•OB•cos∠AOB=AB2即:10+10-20cos∠AOB=8,解得cos∠AOB=35.∴∠AOB的正切值34.
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