+1218 In triangle $ABC,$ $BC = 32,$ $\tan B = \frac{3}{5},$ and $\tan C = \frac{1}{4}.$ Find the area of the triangle.
+129847 Let B = (0,0) C =(32,0)
To find the height of the triangle we can construct two lines using the tangents
y = (3/5)x
And
y = (-1/4) (x -32) = (-1/4)x + 8
To find the x coordinate of their intersection
(3/5()x =(-1/4)x + 8
(3/5 + 1/4)x = 8
[(12 + 5] / 20] x = 8
[17/20] x = 8
x = (8*20] / 17 = 160/17
The height of the triangle (AD) = (3/5)(160/17) = 480 / 85 = 96/17
Area ABC = (1/2) BC * AD = (1/2) 32 * (96/17) = 1536 / 17
