(2)设角A,B,C的对边依次为a,b,c,若c=2,且△ABC是锐角三角形,求a的平方+b的平方的取值范围
(1)tan(A+B)=(tanA+tanB)/(1-tanAtanB)
tan(A+B)(1- tanAtanB)=tanA+tanB=根号3tanAtanB-根号3
tan(A+B)= -根号3,tan(180度-C)=-tanC,tanC=根号3,C=60°
(2)a^2+b^2=(c/sinC)^2 (sin^2 A+sin^2 B)=16/3[ (1-cos2A)/2+(1-cos2B)/2]
=16/3-8/3(cos2A +cos2B)=16/3-16/3cos(A+B)cos(A-B)=16/3+8/3cos(2A-120°)
30°<A <90°, 60°<2A <180°,-60°<2A-120 <60° 1/2<cos(2A-120°)<=1
20/3<a^2+b^2<=8
tan(A+B)(1- tanAtanB)=tanA+tanB=根号3tanAtanB-根号3
tan(A+B)= -根号3,tan(180度-C)=-tanC,tanC=根号3,C=60°
(2)a^2+b^2=(c/sinC)^2 (sin^2 A+sin^2 B)=16/3[ (1-cos2A)/2+(1-cos2B)/2]
=16/3-8/3(cos2A +cos2B)=16/3-16/3cos(A+B)cos(A-B)=16/3+8/3cos(2A-120°)
30°<A <90°, 60°<2A <180°,-60°<2A-120 <60° 1/2<cos(2A-120°)<=1
20/3<a^2+b^2<=8
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第1个回答 2011-04-23
∵√3tanAtanB-tanA-tanB=√3
∴tanA+tanB=-√3(1-tanAtanB)
∵tanA+tanB=tan(A+B)(1-TtanAtanB)
∴tan(A+B)=-√3,
∴tan(180°-C)=-√3
∴tanC=√3
∴C=60°
∴tanA+tanB=-√3(1-tanAtanB)
∵tanA+tanB=tan(A+B)(1-TtanAtanB)
∴tan(A+B)=-√3,
∴tan(180°-C)=-√3
∴tanC=√3
∴C=60°