1樓：匿名使用者
m.n=1
(-1,√3).( cosa, sin a)=1-cosa+√3sin a =1
(√3sin a)^2 = (1+cosa)^22(cosa)^2 + cosa -1 =0(2cosa-1)(cosa+1)=0
cosa = 1/2 or -1 (rejected)a = π/3
(1+sin2b)/(cos^b-sin^b)=-3(1+ sin2b) /cos2b = -3sec2b+ tan2b = -3
(3+tan2b)^2 = (sec2b)^26tan2b+ 9 = 1
tan2b = -4/3
2tanb/(1-(tanb)^2 ) = -4/3-4+4(tanb)^2 = 6tanb
2(tanb)^2 -3tanb-2 =0(2tanb+1)(tanb-2) =0
tanb = 2 or -1/2 (rejected)ie tanb = 2
tanc
= tan(π-a-b)
= tan(2π/3-b)
= (tan2π/3 - tanb)/(1+tanbtan2π/3)=(-√3-2)/(1-2√3)
= ( √3+2)(2√3+1)/ [(2√3-1)(2√3+1)]= (8+5√3)/5
已知a,b,c分別為三角形abc內角a,b,c的對邊，a
解 1 acosc 3asinc b c 0 sinacosc 3sinasinc sinb sinc 0 sinacosc 3sinasinc sinb sinc sin a c sinc sinacosc sinccosa sinc sinc 0 3 sina cosa 1 sin a 30 1...
已知角abc為三角形abc的三內角,其對邊分別為a,b,c,若a 2 b 2 c 2 bc且
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已知角abc為三角形abc的三內角,其對邊分別為a,b,c,若a 2 b 2 c 2 bc且a
1 已知a b c bc 1 所以，bc b c a 由余弦定理有 cosa b c a 2bc bc bc 1 2 所以，a 120 s abc 1 2 bcsina 3 1 2 bc 3 2 3 bc 4 由 1 式得到 12 b c bc b c 12 bc 12 4 8 b c 2bc 8 ...