admin百科知識 2022-02-13 3:58:58 已知三角形ABC的外接圓半逕爲R,且2R(sin^A-sin^C)=(根號2 a-b)sinB.(其中a,b分別已知三角形ABC的外接圓半逕爲R,且2R(sin^A-sin^C)=(根號2 a-b)sinB.(其中a,b分別網友廻答：匿名網友根據正弦定理,a/sinA=b/sinB=c/sinC=2R所以,sinB=b/(2R)2R(sin^A-sin^C)=(√2a-b)*b/(2R)4R^2(sin^A-sin^C)=(√2a-b)*ba^2-c^2=√2ab-b^2c^2=a^2 b^2-√2ab根據餘弦定理,c^2=a^2 b^2-2abcosC得,cosC=√2/2 C=45 db標簽 生活常識_百科知識_各類知識大全»已知三角形ABC的外接圓半逕爲R,且2R(sin^A-sin^C)=(根號2 a-b)sinB.(其中a,b分別
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