math geometry

1. In triangle ABC, AB=12 units and AC=9 units. Point D is on segment BC so that BD:DC=2:1. If AD=6 units, what is the length of segment BC? Express your answer in simplest radical form. 

Let BC = x

cos rule:

\(\begin{array}{|lrcll|} \hline & 6^2 &=& 9^2+(\frac13 x)^2 - 2\cdot 9 \cdot \frac 13 x \cos( C ) \\ & 6^2 &=& 9^2+(\frac13 x)^2 - 6 x \cos( C ) \\ & 6 x \cos( C ) &=& 9^2-6^2+\frac{x^2}{9} \\ (1) & 6 x \cos( C ) &=& 45+\frac{x^2}{9} \\ \hline \end{array}\)

cos rule:

\(\begin{array}{|lrcll|} \hline & 12^2 &=& 9^2+x^2 - 2\cdot 9 \cdot x \cos( C ) \\ & 12^2 &=& 9^2+x^2 - 18 x \cos( C ) \\ & 18 x \cos( C ) &=& 9^2-12^2+x^2 \\ & 18 x \cos( C ) &=& x^2-63 \quad & | \quad : 3 \\ (2) & 6 x \cos( C ) &=& \frac{x^2}{3}-21 \\ \hline \end{array} \)

(2) = (1):

\(\begin{array}{|rcll|} \hline 6 x \cos( C ) = \frac{x^2}{3}-21 &=& 45+\frac{x^2}{9} \\ \frac{x^2}{3}-21 &=& 45+\frac{x^2}{9} \\ x^2(\frac13-\frac19) &=& 66 \\ x^2(\frac{9-3}{27}) &=& 66 \\ x^2(\frac{6}{27}) &=& 66 \\ x^2 &=& 66\cdot (\frac{27}{6}) \\ x^2 &=& 11\cdot 27 \\ x^2 &=& 11\cdot 3^2 \cdot 3 \\ \mathbf{ x} & \mathbf{=} & \mathbf{3\sqrt{33}} \\ \hline \end{array}\)

The length of segment BC is \(\mathbf{3\sqrt{33}}\)

Thanks, heureka....here's my approach....

Using the Law of Cosines we have that

12^2 = (2CD)^2 + 6^2  -  [ 2 * 2CD * 6 ] cos BDA

144 = 4CD^2  + 36 - 24CDcosBDA

108 - 4CD^2  = - 24CDcosBDA

[ 4CD^2 - 108 ] / 24CD  = cos BDA

[CD^2 - 27 ] / 6CD  =cos BDA

Since BDA  and CDA  are supplementary.....-cos BDA  =  cosCDA

Using it again and substituting, we have

9^2 = CD^2 + 6^2  - [ 2 * CD * 6 ] [-cos BDA]

81  = CD^2 + 36 + 12CD  [ CD^2 - 27] / 6CD ]

45 = CD^2 + 2[CD^2 - 27 ]

45 = CD^2 + 2CD^2 - 54

99 = 3CD^2

33 = CD^2

√33 = CD

And  BD  is twice this

So

BD + CD  =  BC  =  √33  + 2√33  =   3√33

Here's a pic :