In triangle ABC, AB = 3, AC = 6, BC = 8, and D lies on BC such that AD bisects ∠BAC. Find cos ∠BAD.
+128407 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
+26367 In triangle ABC, AB = 3, AC = 6, BC = 8, and D lies on BC such that AD bisects ∠BAC. Find cos ∠BAD.
\(\begin{array}{|rcll|} \hline \mathbf{\cos(\angle BAD)} &=& \mathbf{\sqrt{\dfrac{s\cdot (s-BC)}{AB\cdot AC} }} \quad | \quad \mathbf{s = \dfrac{AB+AC+BC}{2}}=\dfrac{3+6+8}{2}=8.5 \\\\ &=& \sqrt{\dfrac{8.5\cdot (8.5-8)}{3\cdot 6} } \\\\ &=& \sqrt{\dfrac{8.5\cdot (0.5)}{18} } \\\\ &=& \sqrt{\dfrac{4.25}{18} } \\\\ &=& \sqrt{\dfrac{2.125}{9} } \\\\ &=& \dfrac{\sqrt{ 2.125 }}{3} \\\\ &=& \dfrac{\sqrt{ 2.125 }}{3} \\\\ \mathbf{\cos(\angle BAD)} &=& \mathbf{ 0.48591265790 } \\ \hline \end{array} \)
