在三角形ABC中,a+c=2b,∠a-∠c=60,求sinB的值?

题目

在三角形ABC中,a+c=2b,∠a-∠c=60,求sinB的值?

答案

在三角形ABC中,
a/sinA = b/sinB = c/sinC
a + c = 2b
所以,
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)
所以,
sin(B/2) = 1/2 *cos30 = 根3 / 4
cos(B/2)=根13/4
sinB =2sin(B/2)cos(B/2)
= 2*根3/4*根13/4
= 根39/8