In a triangle ABC, AC = 2AB and BC/BA = 3/2. H is the foot of the perpendicular from B to AC. Prove that CH/AH = 21/11.
In the triangle ABC, the sides are:
2. BC=1.5* AB
3. AC=2* AB
and CH+AH = AC.
COS(angle A) = AH / AB
COS(angle C) = CH / BC = CH / (1.5* AB)
CH / AH = (1.5* AB)*COS(angle C) / (AB * COS(angle A) )
CH / AH = (1.5)*COS(angle C) / (2 - COS(angle C) )
Hence, CH / AH = 1.90909...= 21 / 11
Is this the correct way to do it?