In triangle $ABC,$ $BC = 32,$ $\tan B = \frac{3}{5},$ and $\tan C = \frac{1}{4}.$ Find the area of the triangle.
Call AD the altitude = x
tan B = AD / BD = 3/5
tan C = AD / CD = 1/4 = 3/12
BD = (5)/(5 + 12) (32) = 160 /17
CD = (12) / (5 +12) (32) = 384/17
So
tan B = 3/5 = AD / BD = x / BD = x / (160/17) → x = (3/5)(160/17) = 96/17 =AD
Check this
tan C = 1/4 = AD /CD = x / CD = x / (384/17) → x = (1/4) (384/17) = 96/17 = AD
[ABC] = (1/2) BC *AD = (1/2) 32 * (96/17) = 1536 / 17