Let $ABC$ be a triangle where $\angle B$ is a right angle. Extend $AC$ up to point $D$ such that $\angle CBD=30^\circ$. If $AB=CD=1$, find $AC$.
My approach to this problem is first to try to find some similarity among the triangle, but did not succeed to find any similar triangle. I then try to use some trigonometry to solve the problem, such as the law of sine, but again, did not succeed.
Any hint of solution to this problem would be appreciated.
On
Let $AC=x$ and $DE$ be an altitude of $\Delta BCD$.
Thus, $\Delta ABC\sim\Delta DEC,$ which gives $$\frac{DE}{1}=\frac{1}{x}$$ or $$DE=\frac{1}{x}$$ and since $\measuredangle DBC=30^{\circ},$ we obtain $$BD=\frac{2}{x}$$ and by the Pythagoras's theorem $$BF=\frac{\sqrt3}{x}.$$
In another hand, by Pythagoras again we obtain: $$BE=BC+CE=\sqrt{x^2-1}+\sqrt{1-\frac{1}{x^2}},$$ which gives $$\sqrt{x^2-1}+\sqrt{1-\frac{1}{x^2}}=\frac{\sqrt3}{x}$$ or $$(1+x)\sqrt{x^2-1}=\sqrt3$$ or $$x^4+2x^3-2x-4=0$$ or $$(x+2)(x^3-2)=0,$$ which gives $$x=\sqrt[3]2.$$
Let $AC=x$. Notice $\angle ACB=\sin^{-1} \frac 1x$ and so $\angle BCD=\pi -\sin^{-1} \frac 1x$. Using the Law of Sines, $$\frac{\sin 30^\circ}{1}=\frac{\sin\left(\pi-\sin^{-1}\frac 1x\right)}{BD} \implies BD=\frac 2x$$ Now use the Law of Cosines on $\triangle BCD$:
$$\cos 30^\circ = \frac{\sqrt 3}{2}=\frac{(\sqrt{x^2-1})^2 + \frac{4}{x^2} -1}{2\sqrt{x^2-1}\cdot \frac{2}{x}} \\ \implies t^4-4t^3-4t+16=0 \\ \implies t=4, 2^{\frac 23}$$ where $t=x^2$ and hence $x=2,2^{\frac 13}$.