Question

在△ABC中,已知cosA=4/5,tanB=5/12,求cosC的值

Answer 1

画个图,很简单,三角形ABC,从C向AB做垂线,交AB于H.cosA = 4/5,设AC = 25k,AH = 20k,

那么CH = 15k.

tanB = 5 / 12,那么 BH = 36k,BC = 39k.

AB = 56k.

用余弦定理：

cosC = (AC^2 + BC^2 - AB^2) / (2 * AC * BC)

= - 33/65