Find $AC$ WITHOUT using Cosine law...

Question

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??

Answer 1

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 $AC=2^{1/3}$.

Answer 2

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 $$x=\sqrt[3]{2}.$$