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Question: In the figure below, AC=AB and AD=BC. Find angle $x$.

enter image description here

My attempt: using a geometric approach, consider the following figure (proportions are not exact).

enter image description here

  1. Using the notation $AC=AB=b$ and $CD=a$, consider point P, in such a way that $AP=b$ and $\angle PAB=60^o$.

  2. Therefore, $\triangle APD\cong \triangle CAB$ (SAS), and $PD=b$.

  3. We can also conclude that $\triangle PAB$ is equilateral and $PB=b$.

  4. Finally, $\triangle DPB$ is isosceles with $\angle DPB=160^o$, so we can conclude that $\angle PBD=10^o$ and $\angle x=10^o$ (as $\angle ABP=60^o$).

Question: is there a trigonometric approach to the problem? I tried using the sinus law but without success. Any hint or full solution using trigonometric methods or other more straightforward approach will be appreciated.

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Let $AB=AC=a$.

Thus, since $$BC=2a\sin50^{\circ},$$ $$DC=AD-AC=BC-AC=2a\sin50^{\circ}-a$$ and $$\measuredangle D=40^{\circ}-x,$$ by law of sines we obtain: $$\frac{\sin{x}}{DC}=\frac{\sin(40^{\circ}-x)}{BC}$$ or $$\frac{\sin{x}}{2\sin50^{\circ}-1}=\frac{\sin(40^{\circ}-x)}{\sin50^{\circ}}$$ or $$\frac{1}{2\sin50^{\circ}-1}=\frac{\sin40^{\circ}\cot{x}-\cos40^{\circ}}{\sin50^{\circ}}$$ or $$\sin40^{\circ}\cot{x}=\frac{2\sin50^{\circ}}{2\sin50^{\circ}-1}+\sin50^{\circ}$$ or $$\cot{x}=\frac{2\sin50^{\circ}+1}{2\sin50^{\circ}-1}\cot40^{\circ}$$ or $$\tan{x}=\frac{2\sin50^{\circ}-1}{2\sin50^{\circ}+1}\tan40^{\circ}$$ and since $$\frac{2\sin50^{\circ}-1}{2\sin50^{\circ}+1}\tan40^{\circ}=\frac{2\cos40^{\circ}-1}{2\cos40^{\circ}+1}\cdot\frac{\sin40^{\circ}}{\cos40^{\circ}}=\frac{\sin80^{\circ}-\sin40^{\circ}}{1+\cos80^{\circ}+\cos40^{\circ}}=$$ $$=\frac{2\sin20^{\circ}\cos60^{\circ}}{1+2\cos20^{\circ}\cos60^{\circ}}=\frac{\sin20^{\circ}}{1+\cos20^{\circ}}=\frac{2\sin10^{\circ}\cos10^{\circ}}{2\cos^210^{\circ}}=\tan10^{\circ},$$ we are done!

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Just one application of Law of Sines is enough. Let me reproduce the diagram below for convenience.

Denote $\angle ACB = \angle ABC$ as $\theta$, which is given as $\theta = 40^{\circ} = \frac{2\pi}9$.

Note that $\angle CDB = \theta - x$ because $\angle ACB$ is the external angle of $\angle DCB$ for $\triangle BCD$. Apply the Law of Sines on $\triangle BCD$, using your notations for the lengths $a$ and $b$: $$\frac{ \sin x }a = \frac{ \sin(\theta - x) }{ 2b \cos \theta } \qquad \text{, where the $2b\cos\theta$ comes from} \triangle ABC ~~\text{being isosceles}$$

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With the given $|AD| = |BC| \implies a+b = 2b \cos\theta~$, namely, $a = b(2 \cos\theta - 1)$, we have $$\sin x \,\frac{ 2\cos\theta }{ 2 \cos\theta - 1} = \sin \theta \cos x - \cos\theta \sin x \\ \sin x \,\cos\theta \left( 1+\frac{ 2 }{ 2 \cos\theta - 1} \right)= \sin \theta \cos x \\ \tan x = \tan\theta \frac{ 2 \cos\theta - 1}{ 2 \cos\theta + 1} $$ Now, looking at this expression, it is tempting to considering something like $x \overset{?}{=} \frac{\theta}4$ by applying twice the half-angle tangent formula. However, this equation holds only for the given $\theta = \frac{2\pi}9$ and one might as well directly consider the analytic properties of $40^{\circ}$ and $10^{\circ}$.

Specifically, for $t\equiv \tan 10^{\circ}$, WolframAlpha tells us that it is a root to the polynomial $p(t)=3t^6 - 27 t^4 + 33 t^2 - 1$ (please scroll down a bit to see that entry).

At the same time, $\displaystyle u \equiv \tan\theta \frac{ 2 \cos\theta - 1}{ 2 \cos\theta + 1}$ is also shown by WolframAlpha to be a root of the same polynomial. Namely $p(u)=0$ just like $p(t) = 0$.

There are six roots (real or complex) of a sextic polynomial. Nonetheless, given our geometric setting (e.g. the known restrictions on the variables) there is no doubt that $$\tan x = u = t = \tan 10^{\circ}$$

Using WolframAlpha is just to out-source the laborious algebraic process. Everything about the tangents being the root of the polynomial are well-established results that can be derived by hand.

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  • $\begingroup$ "...this equation holds only for ... one might as well directly consider the analytic properties ...", oh upon seeing the correct approach by Michael Rozenberg to continue the derivation from this point on, one might as well not! Haha it was a fail on my part. $\endgroup$ – Lee David Chung Lin Oct 13 '18 at 6:29
  • $\begingroup$ Just for the record: the correct relationship between the unknown $x$ and given $\theta = 2\pi/9$ is indeed $x = \theta/4$. The key is to notice that $\theta$ is implicitly a special angle in that $\cos(3\theta/2) = 1/2$. $\endgroup$ – Lee David Chung Lin Oct 13 '18 at 6:44

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