Geometry

Question

0

13

1

+1800

In triangle $ABC,$ $BC = 32,$ $\tan B = \frac{3}{5},$ and $\tan C = \frac{1}{4}.$ Find the area of the triangle.

blackpanther Feb 4, 2025

CPhill · Answer 1 · 2025-02-05T06:32:01

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