根據下列條件,判斷三角形的形狀

[匿名]

2007-02-15 9:21 pm

(1) (cosA)^2+(cosB)^2+(cosC)^2=1

(2) (sinA)^2+(sinB)^2+(sinC)^2<=2

希望有詳細的解題過程,先謝謝了!

回答 (1)

用戶127502

2007-02-16 5:51 am

✔ 最佳答案

Eq (2) is redundant.
By (sinA)^2 + (cosA)^2 = 1,
(cosA)^2+(cosB)^2+(cosC)^2=1
=> (sinA)^2+(sinB)^2+(sinC)^2=2 <=2

(cosA)^2+(cosB)^2+(cosC)^2=1
(cosA)^2+(cosB)^2+(cos(A+B))^2=1 [C=180-(A+B), cos(180-x)=-cos(x)]
(cosA)^2+(cosB)^2=(sin(A+B))^2
(cosA)^2+(cosB)^2=(sinAcosB+cosAsinB)^2
(cosA)^2+(cosB)^2=(sinAcosB)^2+2sinAsinBcosAcosB+(cosAsinB)^2
(cosA)^2(cosB)^2+(cosB)^2(cosA)^2 = 2sinAsinBcosAcosB
(cosA)^2(cosB)^2 = sinAsinBcosAcosB
cosAcosB(cosAcosB-sinAsinB)=0
cosAcosBcos(A+B)=0
cos A = 0 or cos B = 0 or cos(A+B) = 0
=> Right-angled triangle

The manipulation is quite stupid XD

參考: me

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