Question

在三角形ABC中,已知a+c=2b,A-C=60度,求sin(B/2) (a,b,c为边,A,B,C为角)

Answer 1

在三角形ABC中,

a/sinA = b/sinB = c/sinC

a + c = 2b

sinA + sinC = 2sinB

2sin[(A+C)/2] * cos[(A-C)/2] = 2 * 2 * sin(B/2) * cos(B/2)

sin[(A+C)/2] * cos30 = 2 * sin(B/2) * cos(B/2)

cos[90 - (A+C)/2] * cos30 = 2 * sin(B/2) * cos(B/2)

cos(B/2) * cos30 = 2 * sin(B/2) * cos(B/2)

cos30 = 2 * sin(B/2)

sin(B/2) = 根3 / 4