(05复旦)已知三角形$Delta ABC$满足$ an A: an B: an C=1:2:3$,求$dfrac{AC}{AB}$____
解答:设$x=tan A$,利用恒等式$ an A an B an C= an A+ an B+ an C$
得$6x^3=6x$故$x=1, an B=2x=2,tan C=3x=3$
进而得$sin B=dfrac{2}{sqrt{5}},sin C=dfrac{3}{sqrt{10}}$故$dfrac{AC}{AB}=dfrac{sin B}{sin C}=dfrac{2sqrt{2}}{3}$
或者利用切割化弦,都变成边也可得到答案.