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

 
 Feb 4, 2025
 #1
avatar+130313 
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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

 

 

cool cool cool

 Feb 5, 2025

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