In a triangle ABC, the median CM is perpendicular to angular bisector AL, M being on AB and L being on BC. Also CM/AL = n. Find cosA in terms of n. 1) (9+4n²)/(9-4n²)2) (9-4n²)/(9+4n²)3) (5-4n²)/(5+4n²)4) (9-n²)/(9+n²)

In a triangle ABC, the median CM is perpendicular to angular bisector

1.	In a triangle ABC, the median CM is perpendicular to angular bisector AL, M being on AB and L being on BC. Also CM/AL = n. Find cosA in terms of n. 1) (9+4n²)/(9-4n²)2) (9-4n²)/(9+4n²)3) (5-4n²)/(5+4n²)4) (9-n²)/(9+n²)
Answer» In a triangle ABC, the median CM is perpendicular to angular bisector AL, M being on AB and L being on BC. Also CM/AL = n. Find cosA in terms of n. 1) (9+4n²)/(9-4n²) 2) (9-4n²)/(9+4n²) 3) (5-4n²)/(5+4n²) 4) (9-n²)/(9+n²)