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[三角][謝謝老師]
版主: thepiano
Re: [三角]
△ABC = (1/2)absinC
= (1/2)ab√[1 - (cosC)^2]
= (1/2)ab√[1 - (a^2 + b^2 - c^2)^2 / (2ab)^2]
= (1/4)√[4a^2b^2 - (a^2 + b^2 - c^2)^2]
≦ (1/4)√[(a^2 + b^2)^2 - (a^2 + b^2 - c^2)^2]
= (1/4)√[(8 - 2c^2)^2 - (8 - 2c^2 - c^2)^2]
= (1/4)√(-5c^4 + 16c^2)
......
可求出答案為 (2/5)√5
= (1/2)ab√[1 - (cosC)^2]
= (1/2)ab√[1 - (a^2 + b^2 - c^2)^2 / (2ab)^2]
= (1/4)√[4a^2b^2 - (a^2 + b^2 - c^2)^2]
≦ (1/4)√[(a^2 + b^2)^2 - (a^2 + b^2 - c^2)^2]
= (1/4)√[(8 - 2c^2)^2 - (8 - 2c^2 - c^2)^2]
= (1/4)√(-5c^4 + 16c^2)
......
可求出答案為 (2/5)√5