在锐角△ABC中,a,b,c分别是∠A,∠B,∠C的对边,切tanA=2tanB,sin(A-B)=根号3/6

题目

在锐角△ABC中,a,b,c分别是∠A,∠B,∠C的对边,切tanA=2tanB,sin(A-B)=根号3/6
1.求∠C的大小.
2.若3ab=25-c^2,△ABC面积为 25根号3/16,判断△ABC的形状.

答案

tanA = 2tanB
sinA/cosA = 2sinB/cosB
sinAcosB - 2cosAsinB = 0
sin(A-B) = sinAcosB - cosAsinB = √3 /6
解得 cosAsinB = √3/6 ,sinAsinB = √3/3
sinC = sin(π - C) = sin(A+B)
=sinAcosB+cosAsinB
= √3 /2
∠C = π/3
S = 1/2 absinC = √3/4 ab = 25√3 /16
∴ ab = 25/4
3ab = 25 -c²
c² = 25 - 3ab = 25/4
c = 5/2
由余弦定理,c² = a²+b² - 2abcosC = a²+b² - ab = a²+b² - 25/4
a²+b² = c²+ 25/4 = 25/2
(a+b)² = a²+b²+2ab = 25/2 + 2*25/4 = 25
a+b = 5
又 ab =25/4
∴a=b= 5/2
a=b=c
△ABC是等边三角形