In triangle $ABC,$if $AC=8,BC=7$ and $D$ lies between $A$ and $B$ such that $AD=2,BD=4$,then find length $CD.$

Using cosine law,i found $C=\arccos(\frac{11}{16})$

Now $\angle ACD=\frac{1}{3}C=\frac{1}{3}\arccos(\frac{11}{16})$
In triangle $ACD,$

$\cos\angle ACD=\cos(\frac{1}{3}\arccos(\frac{11}{16}))=\frac{8^2+CD^2-2^2}{2\times 8\times CD}$
I am stuck here.Is my method wrong?Is there some other more efficient method to solve this problem.

  • $\begingroup$ It is not true that $\angle ACD=\frac13\angle C$. $\endgroup$ – Ángel Mario Gallegos Apr 10 '16 at 16:48

Using the Cosine Law we have \begin{align*} |CD|^2&=|AC|^2+|AD|^2-2|AC|\cdot |AD|\cos \angle CAB\\[5pt] &=|AC|^2+|AD|^2-2|AC|\cdot |AD|\cdot\left(\frac{|AC|^2+|AB|^2-|BC|^2}{2|AC|\cdot|AB|}\right)\\[5pt] &=|AC|^2+|AD|^2-|AD|\cdot\left(\frac{|AC|^2+|AB|^2-|BC|^2}{|AB|}\right)\\[5pt] &=8^2+2^2-2\cdot\left(\frac{8^2+6^2-7^2}{6}\right)\\[5pt] &=64+4-2\cdot\frac{51}6\\ &=51 \end{align*} Then $$\boxed{\color{blue}{|CD|=\sqrt{51}}}$$ enter image description here

  • $\begingroup$ Why |AB| = 8 in the third step. It should be 6 and the answer should be $\sqrt{51}$ $\endgroup$ – Zack Ni Apr 12 '16 at 8:42
  • $\begingroup$ @ZackNi: Yes, you are right. See the changes I have done in my post. $\endgroup$ – Ángel Mario Gallegos Apr 12 '16 at 14:18

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