We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
76
2
avatar

In triangle ABC, AB = 3, AC = 6, BC = 8, and D lies on BC such that AD bisects ∠BAC. Find cos ∠BAD.

 Sep 18, 2019
 #1
avatar+104876 
+1

See the following image :

 

 

Since AD bisects BAC   then 

BA /CA  =  BD/DC

3 / 6   =  BD /DC

1 /2   =  BD/DC

DC = 2BD

BC = BD + DC

BC = BD + 2BD

BC = 3BD

So  BD =  (1/3)BC =  (1/3)(8)  = 8/3

And  DC = 2BD = 2(8/3)  =  16/3

 

 

So....using the Law of Cosines we have

 

(8/3)^2  =  AD^2 + 3*2 - 2(AD)(3) cosBAD   (1)

(16/3)^2  = AD^2 +6^2  - 2(AD)(6) cosDAC   (2)

 

And since AD  bisects BAC....then cos  BAD  =  cos DAC

 

So.....

 

(64/9) - 9  = AD^2 - 6(AD)cosBAD     (3)

(256/9) - 36  = AD^2 - 12AD cosBAD  (4)

 

-17/9  = AD^2 - 6(AD) cosBAD   (5)

-68/9  = AD^2 - 12(AD) cosBAD    (6)   

 

Multiply (5)  by -2

 

34/9  = -2AD^2 + 12(AD)cosBAD   (7)

-68/9  =  AD^2  - 12(AD)cosBAD    (8)       add (7) and (8)

 

-34/9 = -AD^2   

34/9  = AD^2

√34/3 = AD

 

Therefore....using (5)  and subbing for AD....  we have that

 

-17/9 = 34/9 - 6(√34/3)cosBAD

 

-51/9  = - 2√34cosBAD

 

cos BAD  =  (51) / ( 9 * 2√34)

 

cos BAD  = (51) (18√34)

 

cos BAD  =  17/ (6√34)  =  (17√34) /  (6 * 34)  =  √34 / 12

 

 

 

cool cool  cool

 Sep 19, 2019
 #2
avatar+23317 
+2

In triangle ABC, AB = 3, AC = 6, BC = 8, and D lies on BC such that AD bisects ∠BAC. Find cos ∠BAD.

 

\(\begin{array}{|rcll|} \hline \mathbf{\cos(\angle BAD)} &=& \mathbf{\sqrt{\dfrac{s\cdot (s-BC)}{AB\cdot AC} }} \quad | \quad \mathbf{s = \dfrac{AB+AC+BC}{2}}=\dfrac{3+6+8}{2}=8.5 \\\\ &=& \sqrt{\dfrac{8.5\cdot (8.5-8)}{3\cdot 6} } \\\\ &=& \sqrt{\dfrac{8.5\cdot (0.5)}{18} } \\\\ &=& \sqrt{\dfrac{4.25}{18} } \\\\ &=& \sqrt{\dfrac{2.125}{9} } \\\\ &=& \dfrac{\sqrt{ 2.125 }}{3} \\\\ &=& \dfrac{\sqrt{ 2.125 }}{3} \\\\ \mathbf{\cos(\angle BAD)} &=& \mathbf{ 0.48591265790 } \\ \hline \end{array} \)

 

laugh

 Sep 19, 2019

33 Online Users

avatar
avatar
avatar