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# Precal Question

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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
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See the following image : 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

So.....

Multiply (5)  by -2

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

cos BAD  =  (51) / ( 9 * 2√34)

cos BAD  =  17/ (6√34)  =  (17√34) /  (6 * 34)  =  √34 / 12   Sep 19, 2019
#2
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In triangle ABC, AB = 3, AC = 6, BC = 8, and D lies on BC such that AD bisects ∠BAC. Find cos ∠BAD.

$$\begin{array}{|rcll|} \hline \mathbf{\cos(\angle BAD)} &=& \mathbf{\sqrt{\dfrac{s\cdot (s-BC)}{AB\cdot AC} }} \quad | \quad \mathbf{s = \dfrac{AB+AC+BC}{2}}=\dfrac{3+6+8}{2}=8.5 \\\\ &=& \sqrt{\dfrac{8.5\cdot (8.5-8)}{3\cdot 6} } \\\\ &=& \sqrt{\dfrac{8.5\cdot (0.5)}{18} } \\\\ &=& \sqrt{\dfrac{4.25}{18} } \\\\ &=& \sqrt{\dfrac{2.125}{9} } \\\\ &=& \dfrac{\sqrt{ 2.125 }}{3} \\\\ &=& \dfrac{\sqrt{ 2.125 }}{3} \\\\ \mathbf{\cos(\angle BAD)} &=& \mathbf{ 0.48591265790 } \\ \hline \end{array}$$ Sep 19, 2019