Triangle angle question

Question

I need help about a triangle angle question.

Answer 1

Flip the whole image through the symmetry median of the $20-80-80$ triangle. The outer part forms a large equilateral triangle.

Construct a smaller equilateral triangle outside the $80-80$ edge.

Find the reflective and rotational symmetry, and check that $\alpha$ is half of $20^\circ$.

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Edit: changed diagram to name points

Answer 2

There is typically a clever trick to solving these problems, but if you do not know the trick you can always just throw coordinates at them.

Let $x$ be the length of the three equal-sized lines, and let the point at the bottom left be at ($0$,$0$).

The point at the top is then at ($x\cos(40^{\circ})$,$x\sin(40^{\circ})$).

The point at the bottom right is at ($2x\cos(20^{\circ})$,$0$)

so

$$\tan(20^{\circ}+\alpha) = \frac{x\sin(40^{\circ})}{2x\cos(20^{\circ}) - x\cos(40^{\circ})}$$

$$\tan(20^{\circ}+\alpha) = \frac{\sin(40^{\circ})}{2\cos(20^{\circ}) - \cos(40^{\circ})}$$

$$20^{\circ}+\alpha = \arctan\left( \frac{\sin(40^{\circ})}{2\cos(20^{\circ}) - \cos(40^{\circ})} \right)$$

The r.h.s. is $30^{\circ}$

$$20^{\circ}+\alpha = 30^{\circ}$$

$$\alpha = 10^{\circ}$$