Question

设三角形的内角abc所对的边分别为abc,且a-c=1,b=2,cosB=3/5(1)求a+c的值;(2)求cos(A-B)的值

要解题过程

Answer 1

(1)
b^2 = a^2+c^2-2ac.cosB
= (a-c)^2+2ac-2ac.cosB
4=1+2ac(1-3/5)
=1+ (4/5)ac
ac = 15/4

(a+c)^2 = (a-c)^2 +4ac
=1 + 15
a+c =4

(2)
cosB= 3/5 => sinB = 4/5

a+c = 4 (1)
a-c= 1 (2)
(1)+(2)
a = 5/2
a/sinA = b/sinB
sinA = asinB/b
=(5/2) (4/5) / ( 2)
= 1
=> cosA = 0

cos(A-B)
=cosAcosB + sinAsinB
=0+ 1(4/5)
=5/5