△ABC从顶点拉出一条线垂直BC，给了∠B=60度, ∠A=45度，AB=5，问AC=?

yunduanxw

这题怎么做啊？我要详细的步骤哎！

谢谢啦！

sdcar2010

As shown in the graph, AB = 5, BD = 2.5, AD = 2.5*3^(1/2) ∠B = 60, ∠BAD = 30, ∠DAC=15, ∠ADB=90.

Add two extra lines: AE bisects ∠BAD; EF is perpendicular to AB. Therefore AE = AC; AF = AD; EF=DE.

Let DE = x.

BE/EF = 2/[3^(1/2) ]; so BE = [2x]/[3^(1/2) ]; so BD = BE + DE = [2x]/[3^(1/2) ] + x = 2.5; so x = 5*3^(1/2) - 7.5

AC = AE = [DE^2 + AD^2]^(1/2) = [x^2 + 75/4]^(1/2) = (150-75*3^(1/2)]^(1/2)

wangxiao89

用正弦定理啊

sdcar2010

Then the tester needs to know

sin(75) = sin (30+45) = sin30 * cos 45 + cos 30 * sin 45

夏子之酒

还是解不出来阿BD = BE + DE = [2x]/[3^(1/2) ] + x = 2.5; so x = 5*3^(1/2) - 7.5   最后x 不知道怎么推出来的。。。别拍我。。。。x=5*[3^(1/2) ]/2*[3^(1/2) ]+4  .......

爱之兮3344

我的答案是(15√2-5√6)/2，值大概是4.483，不知道有没有相同的童鞋~~