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In triangle $ABC,$ $AB = 3,$ $AC = 5,$ $BC = 7,$ and $D$ lies on $\overline{BC}$ such that $\overline{AD}$ bisects $\angle BAC.$ Find $\cos \angle BAD.$

 Jun 19, 2024
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avatar+129725 
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                      A

              

      3                                5

 

B                          D                      C

                      7

 

Law of Cosines

 

BD^2 = AB^2 + AC^2 - 2(AB * AC) cos BAC

7^2  = 3^2  + 5^2  - 2 (3 * 5) cos BAC

cos BAC  = [ 7^2 - 3^2 - 5^2 ] / [ -2 (3 * 5) ]  =  -1/2

 

arccos (-1/2)  =  measure of BAC  =  120°

 

Since AD is an angle bisector

 

BAD  = 60°

 

cos (BAD)   = cos (60°)  =  1/2

 

cool cool cool                

 Jun 20, 2024

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