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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)
人气:387 ℃ 时间:2019-10-19 01:07:38
解答
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
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