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In right triangle $ABC$, we have $\angle BAC = 90^\circ$ and $D$ is on $\overline{AC}$ such that $\overline{BD}$ bisects $\angle ABC$. If $AB = 12$ and $BC = 15$, then what is $\cos \angle BDC$?

 Mar 24, 2020
edited by Guest  Mar 24, 2020
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See the following pic :

 

 

 

Let's  find BC =  √ [AB^2  + AC^2) √  (12^2 + 15^2) =  3√ 41

 

Because  ABC   is bisected, we  have the  folllwing relationship :

 

AD / DC  =  AB/ BC   so

 

AD/DC  = 12/ [3√ 41]  =   4/√ 41

 

So.....AD  =   15  ( 4 / [4 +√ 41])  =    60/ [ 4 + √ 41]

 

And  we  can find  BD  as  √   [ 12^2  + (60/ [ 4 + √ 41])^2 ]

 

Then  triangle BDA  is right

 

So  cos  BDA =     AD / BD  =   (60/ [ 4 + √ 41 ] ) /    √   [ 12^2  + (60/ [ 4 + √ 41])^2 ]  =

 

Using WolframAlpha to simplify this (mess) we get that 

 

cos BDA  =   5 / √  [82  + 8√ 41]

 

And  because BDC  is supplemental to this we have  that   

 

cos BDC  =   - 5 /  √ [82 + 8√ 41]

 

 

cool cool cool

 Mar 24, 2020

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