1.三角形ABC中 已知lgtanB是lgtanA ,lgtanC的算术平均值,求角B的取值范围?2.已知sina+sinb+sinc=0并且cosa+cosb+cosc=0,则cos(a-b)的值是多少3.三角形ABC中,sinA+cosA=2/3,则三角形ABC的形状为?
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![1.三角形ABC中 已知lgtanB是lgtanA ,lgtanC的算术平均值,求角B的取值范围?2.已知sina+sinb+sinc=0并且cosa+cosb+cosc=0,则cos(a-b)的值是多少3.三角形ABC中,sinA+cosA=2/3,则三角形ABC的形状为?](/uploads/image/z/5183259-51-9.jpg?t=1.%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD+%E5%B7%B2%E7%9F%A5lgtanB%E6%98%AFlgtanA+%2ClgtanC%E7%9A%84%E7%AE%97%E6%9C%AF%E5%B9%B3%E5%9D%87%E5%80%BC%2C%E6%B1%82%E8%A7%92B%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4%3F2.%E5%B7%B2%E7%9F%A5sina%2Bsinb%2Bsinc%3D0%E5%B9%B6%E4%B8%94cosa%2Bcosb%2Bcosc%3D0%2C%E5%88%99cos%28a-b%29%E7%9A%84%E5%80%BC%E6%98%AF%E5%A4%9A%E5%B0%913.%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2CsinA%2BcosA%3D2%2F3%2C%E5%88%99%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E7%9A%84%E5%BD%A2%E7%8A%B6%E4%B8%BA%3F)
1.三角形ABC中 已知lgtanB是lgtanA ,lgtanC的算术平均值,求角B的取值范围?2.已知sina+sinb+sinc=0并且cosa+cosb+cosc=0,则cos(a-b)的值是多少3.三角形ABC中,sinA+cosA=2/3,则三角形ABC的形状为?
1.三角形ABC中 已知lgtanB是lgtanA ,lgtanC的算术平均值,求角B的取值范围?
2.已知sina+sinb+sinc=0并且cosa+cosb+cosc=0,则cos(a-b)的值是多少
3.三角形ABC中,sinA+cosA=2/3,则三角形ABC的形状为?
1.三角形ABC中 已知lgtanB是lgtanA ,lgtanC的算术平均值,求角B的取值范围?2.已知sina+sinb+sinc=0并且cosa+cosb+cosc=0,则cos(a-b)的值是多少3.三角形ABC中,sinA+cosA=2/3,则三角形ABC的形状为?
1.
lgtanB是lgtanA ,lgtanC的算术平均值 ,故:
lgtanA + lgtanC = 2lgtanB = lg(tanB)^2 = lg[(tanA)·(tanC)] ,
所以 ,(tanB)^2 = (tanA)(tanC) ,由于B = π - (A+B),故:
-tanB = tan(A+C) = [tanA + tanC]/[1 - (tanA)(tanC)] =
[tanA + tanC]/[1 - (tanB)^2] ,故(tanB)[(tanB)^2 -1]=[tanA + tanC],
因为tanA、tanB、tanC都是对数的真数 ,故均大于零 ,从而:
tanA + tanC 》2(tanA·tanC)^(1/2) = 2tanB ,代入即得:
(tanB)[(tanB)^2 -1] 》2tanB ,即(tanB)^2 》3 ,tanB 》根3 ,
B 》π/3 ,由于tanB > 0使得cosB > 0 ,故B为锐角 ,B的范围是:
【π/3 ,π/2),当且仅当A=B=C=π/3取等号.
2.
由题意,sinc = -(sina + sinb) ,cosc = -(cosa + cosb) ,
两式分别平方:(sinc)^2 = (sina + sinb)^2
(cosc)^2 = (cosa + cosb)^2 ,
而 1 = (sinc)^2+(cosc)^2
= (sina)^2 + (cosa)^2 + (sinb)^2 + (cosb)^2
+ 2cosa·cosb + 2sina·sinb
= (sina)^2 + (cosa)^2 + (sinb)^2 + (cosb)^2 + 2cos(a-b)
= 2 + 2cos(a-b) ,所以cos(a-b) = -1/2
3.
显然在三角形ABC中 ,sinA > 0 ,
sinA+cosA=2/3 ,所以cosA = 2/3 - sinA ,代入同角关系式解得:
sinA = (1/3) + (根14)/6 ,(负值舍),
cosA = 2/3 - sinA = (1/3) - (根14)/6 < 0 ,所以A > 90°,
三角形ABC是以A为钝角的钝角三角形.