# Finding an angle in equilateral triangle

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

Here is my attempt:

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?

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, $$\cot10^{\circ}\cdot\frac{1-2\sin10^{\circ}}{1+2\sin10^{\circ}}=\cot10^{\circ}\cdot\frac{\sin30^{\circ}-\sin10^{\circ}}{\sin30^{\circ}+\sin10^{\circ}}=$$ $$=\cot10^{\circ}\cdot\frac{2\sin10^{\circ}\cos20^{\circ}}{2\sin20^{\circ}\cos10^{\circ}}=\cot20^{\circ}=\tan70^{\circ}.$$