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Given $\triangle ABC$ ($AC=AB$). $X$ - the point on side $AC$ such as $AX=BC$. $\angle A = 20^0$. Find $\angle XBC$.

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Here is my attempt:

figure triangle

Let side $AB = a$, then side $BC = 2a \sin10^0$. Construct $B_1X \parallel BC$.

Similar triangles $\triangle BAC \sim \triangle B_1AX$ gives $B_1X=4a\sin^2 10^0$.

$XK \perp BC$. From $\triangle CXK$: $XK=XC\cdot \cos 10^0=(a-2a\sin 10^0)\cos10^0$.

$BK= \frac{B_1X+BC}{2}=2a\sin^2 10^0+a\sin 10^0$.

$\tan XBK = \frac{KX}{BK}= \frac{\cos 10^0(1-2\sin 10^0)}{\sin 10^0(1+2\sin 10^0)}= \cot 10^0 \cdot \frac{(1-2\sin 10^0)}{(1+2\sin 10^0)}$

Then I find perfect solution of this problems by @Seyed in this post Find $x$ angle in triangle.

That's why I have a question: is $\tan 70^0$ equal $\cot 10^0 \cdot \frac{(1-2\sin 10^0)}{(1+2\sin 10^0)}$ or I have a mistake in my attempt?

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Let $Y\in CX$,$Z\in AB$ and $X'\in AY$ such that $BY=ZY=ZX'.$

Thus, $$\measuredangle ZX'Y=\measuredangle X'YZ=180^{\circ}-\measuredangle ZYB-\measuredangle BYC=180^{\circ}-60^{\circ}-80^{\circ}=40^{\circ},$$ which gives $$\measuredangle AZX'=\measuredangle ZX'Y-\measuredangle A=40^{\circ}-20^{\circ}=20^{\circ},$$ which says $$AX'=ZX'=BC,$$ which gives $$X'\equiv X.$$ Id est, $$\measuredangle XBC=\measuredangle ABC-\measuredangle XBZ=80^{\circ}-\frac{1}{2}\cdot20^{\circ}=70^{\circ}.$$

By the way, you are right: $$\tan70^{\circ}=\cot10^{\circ}\cdot\frac{1-2\sin10^{\circ}}{1+2\sin10^{\circ}}.$$ Indeed, $$\cot10^{\circ}\cdot\frac{1-2\sin10^{\circ}}{1+2\sin10^{\circ}}=\cot10^{\circ}\cdot\frac{\sin30^{\circ}-\sin10^{\circ}}{\sin30^{\circ}+\sin10^{\circ}}=$$ $$=\cot10^{\circ}\cdot\frac{2\sin10^{\circ}\cos20^{\circ}}{2\sin20^{\circ}\cos10^{\circ}}=\cot20^{\circ}=\tan70^{\circ}.$$

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