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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 \(\cos \angle BDC\)?

 Jun 13, 2019
 #1
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B

 

12                        15

 

 

A                    D                        C

 

We have a 9 - 12 - 15   right triangle

 

So   AC   = 9

 

Let  AD = x    and DC = 9 -x

 

And since BD bisects angle ABC,we have the following relationship

 

AB/ BC  =  AD / DC

 

12/ 15 =   x / (9 - x)    cross-multiply

 

12 ( 9 - x)  =  15x

 

108 - 12x  = 15x

 

108  = 27x

 

x = 4  =   AD

 

And  BD  = √[ AD^2  +   BA^2]  =  √[ 4^2 + 12^2 ]  = √160  = 4√10

 

And since  angles  BDC  and ADB  are supplementary

 

cos (BDA)  = -cos (BDC)

 

- cos(BDA) = cos(BDC)

 

And cos  (BDA)  = AD/ BD   =    4 / [ 4√10]  =  1 / √10  =  √ 10  /  10

 

So   cos (BDC) = - cos (BDA) =  - √10/ 10

 

cool cool cool

 Jun 13, 2019

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