题目

b2+c2-bc=a2 b/c=1/2+根号3 求tanB

提问时间：2020-07-24

答案

在三角形ABC中,
2bc*cosA = b^2 + c^2 - a^2 = bc,
cosA = 1/2.
A = 60度.
B + C = 180度 - 60度 = 120度
C = 120度 - B.
sinC = sin(120度 - B) = sin(120度)cosB - cos(120度)sinB
= [3^(1/2)cosB + sinB]/2,
sinC/sinB = [3^(1/2)/tanB + 1]/2,
1/2 + 3^(1/2) = b/c = sinB/sinC = 2/[3^(1/2)/tanB + 1],
2 = [3^(1/2)/tanB + 1][1/2 + 3^(1/2)]
= 3^(1/2)/(2tanB) + 1/2 + 3/tanB + 3^(1/2)
3/2 - 3^(1/2) = [3^(1/2) + 6]/(2tanB)
tanB = [3^(1/2) + 6]/[3 - 2*3^(1/2)]
= [3^(1/2) + 6][3 + 2*3^(1/2)]/[9 - 12]
= [3*3^(1/2) + 18 + 6 + 12*3^(1/2)]/(-3)
= -[5*3^(1/2) + 8]