I would like to ask you for help for the following problem:
The sides of a $\triangle ABC$ are $BC=2,AB=5,AC=\sqrt{22}$ and point $D$ lies on $AC$, such that $BD=3$. Find $CD$.
$CD=\dfrac{\sqrt{22}}{2}$
I have just studied the Cosine rule and I think I am supposed to use it. I don't see how, though. To find $CD$ we can use the Cosine rule in $\triangle BCD:CD^2=BC^2+BD^2-2.BC.BD.\cos\measuredangle CBD$ but I don't see how to find $\cos\measuredangle CBD$. I also found $\cos\gamma=\dfrac{a^2+b^2-c^2}{2ab}=\dfrac{1}{4\sqrt{22}}=\dfrac{\sqrt{22}}{88}.$
Best Answer
First, use the cosine rule on triangle ACB to find $\measuredangle ACB$.
Then use the cosine rule on triangle BCD to find the length of $CD$.