Find $AC$ WITHOUT using Cosine law... 
In the following figure find side $AC$ WITHOUT using Cosine law:
  

There is a simple solution involving Cosine law,but is it possible to calculate $AC$ without using it?? 
 A: Let $DK$ be an altitude of $\Delta ABD$, $AC=x$ and $BD=y$.
Thus, $$DK=\frac{y\sqrt3}{2},$$ 
$$BK=\frac{y}{2}$$ and
$$BC=\sqrt{x^2-1}.$$
Thus, since $$\frac{AB}{BK}=\frac{AC}{CD}$$ and
$$\frac{BC}{DK}=\frac{AC}{AD},$$ we got the following system:
$$\frac{1}{\frac{y}{2}}=\frac{x}{1}$$ and
$$\frac{\sqrt{x^2-1}}{\frac{y\sqrt3}{2}}=\frac{x}{x+1}.$$
Thus, $$\frac{\sqrt{x^2-1}}{\frac{\sqrt3}{x}}=\frac{x}{x+1}$$ or
$$\sqrt{x^2-1}(x+1)=\sqrt3$$ or
$$x^4+2x^3-2x-4=0$$ or
$$(x+2)(x^3-2)=0$$ or
$$x=\sqrt[3]{2}.$$
A: Let $BC=x$. Then $AC=\sqrt{1+x^2}$. Use the law of sines in triangle BCD to obtain
$$\frac{x}{\sin\angle D}=\frac{1}{\sin 30^\circ}=2.$$
Notice that $\angle ABD=120^\circ$, whose supplementary angle is $60^\circ$. So we have
$$ AB\times\sin 60^\circ=AD\times\sin\angle D.$$
We may plug in numbers to obtain
$$(1+\sqrt{1+x^2})\,\frac{x}{2}=\frac{\sqrt{3}}{2}.$$
Here's the tricky part. In terms of $y=\sqrt{1+x^2}$, the equation becomes
\begin{align}
(1+y)\sqrt{y^2-1}&=\sqrt{3},\\
y^4+2y^3-2y-1&=3,\\
(y^3-2)(y+2)&=0.
\end{align}
The answer is $AC=2^{1/3}$.
