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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 $\cos \angle BAD.$

 Jan 23, 2025
 #1
avatar+130317 
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                  A

    

          3               5

 

B               D                    C

                  7

 

Law of Cosines

 

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

 

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

 

BAC  = 120°

 

Since BAD is  a bisector

 

BAD = 60°

 

cos (BAD)  = 1 / 2

 

 

cool cool cool

 Jan 23, 2025

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