Question

三角形ABC中，内角A,B,C成等差数列，其对边A,B,C满足2b方=3ac 求A

急用

Answer 1

解：∵在三角形ABC中，内角A,B,C成等差数列
∴　A＋B＋C＝　　180°　　2B = A + C 　∴ A + C = 2B= 120 ° B =60° sinB = √3/2 sin2B = 3/4 根据正弦定理 有
sin2B = 3/4
a/sinA = b/sinB = c/sinC a =( b/sinB) ×sinA c = ( b/sinB) ×sinC
由已知 2b2 = 3ac = 3 ×( b/sinB) ×sinA×( b/sinB) ×sinC
2 = 3 ×sinA×sinC/ sin2B sinA×sinC = 2×sin2B/3 = 2×3/4/3 = 1/2
A + C = 2B= 120 °
∴ cos (A+C) =cos120 °= -1/2
cos (A+C) =cosAcosC – sinA sinC -1/2 = cosAcosC -1/2 cosAcosC = 0
∵cosAcosC = 0 ∴ A C 至少有一个是90° A == 90°