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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 26, 2024
 #1
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+1

              A

       3           5

B           D            C 

              7

 

Law of Cosines

 

7^2 = 3^2 + 5^2  - 2(3*5)cos ( BAC)

 

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

 

arccos (-1/2)  = BAC =  120°

 

So  cos (BAD)  =   cos (120 / 2) =  cos 60 =  1/2

 

cool cool cool

 Jun 26, 2024

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