Question

在三角形ABC AD垂直于BC 垂足为D 且BD:DC:DF=2:3:6 求角BAC的度数

Answer 1

令,BD:DC:DA=2:3:6=2m:3m:6m,

则有,

BD=2m,CD=3m,AD=6m.

AB^2=AD^2+BD^2=40m^2,

AC^2=AD^2+CD^2=45m^2.

cos角BAC=(AB^2+AC^2-BC^2)/(2AB*AC)=√2/2=cos45,

则,角BAC=45度.