在三角形abc中,b=60° ab=8点d在bc上且cd=2 cos角adc=1⼀7求sinba

2025-05-22 10:25:51
推荐回答(3个)
回答(1):

简单分析一下,答案如图所示

回答(2):

过A作底边的垂线,运用三角函数以及和差公式,可以求出sin∠BAD=√3*3/14
BD=3,AC=7

回答(3):

cos ADC=1/7, cosADB = -1/7
sin BAD = sin(180-B-ADB) = sin(B+ADB) = sinBcosADB+cosBsinADB
= sqrt(3)/2 * (-1/7) + 1/2 4sqrt(3)/7 = 3sqrt(3)/14
sinADB / AB = sinBAD/BD
BD = AB * sinBAD/sinADB = 8 * 3sqrt(3)/14 / (4sqrt(3)/7) = 3
BC = 5
AC^2 = AB^2 + BC^2 - 2AB*BC cosB = 8^2 + 5^2 - 8*5 = 49
AC = 7