一道数学题如图，在△ABC中，∠
假设 AF= x , 则 AC =2x
因∠FAC =120,由余弦定理得:
FC^2 = x^2+(2x)^2-2x*2x*cos 120 = 5x^2+2x^2= 7x^2
FC = rt(7)*x
再由正弦定理可以求出 ∠FCA的正弦:
FC/sin 120 = x/sin ∠FCA ==> sin∠FCA = x * rt(3)/2 /(rt(7)* x) = rt(21)/14
进而求出其余弦供后面使用:
cos ∠FCA = rt((1- sin∠FCA^2)) = rt(175)/14 = 5rt(7)/14
同时因为
∠BAC=120及AB=AC ==> ∠ABC =...全部
假设 AF= x , 则 AC =2x
因∠FAC =120,由余弦定理得:
FC^2 = x^2+(2x)^2-2x*2x*cos 120 = 5x^2+2x^2= 7x^2
FC = rt(7)*x
再由正弦定理可以求出 ∠FCA的正弦:
FC/sin 120 = x/sin ∠FCA ==> sin∠FCA = x * rt(3)/2 /(rt(7)* x) = rt(21)/14
进而求出其余弦供后面使用:
cos ∠FCA = rt((1- sin∠FCA^2)) = rt(175)/14 = 5rt(7)/14
同时因为
∠BAC=120及AB=AC ==> ∠ABC =30 加上BD⊥BC =>∠ABD=120 = ∠FAC
BD= AF, BA= AC, ∠ABD = ∠CAF ==> △ABD = △CAF ==> ∠BAD=∠ACF= ∠FCH
∠BAD=∠FCH,∠FGA=∠HGC => △AFG ~ △CHG ==>(添加辅助线FH)△FGH~△AGC ==> ∠FHA=∠FCA = ∠BAD
==> AF = FH =x
在△FHA中再使用正弦定理
x/sin∠FAD = 5/sin(180-2∠FAD)
==> x/sin∠FAD = 5/sin2∠FAD
==> x/sin∠FAD = 5/(2*sin∠FAD*cos∠FAD)两边同乘以 sin∠FAD
==> x = 5/(2cos∠FAD) = 5/(2cos∠FCA)
= 5/(2*5rt(7)/14)= 14/2*rt(7) = 7/rt(7) = rt(7)
最后答案是根号7 ,约等于 2。
65。收起
