Triangle angle question

Question

I need help about a triangle angle question.

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score 3 · Answer 1 · edited Jun 12 '20 at 10:38

3

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.

construction

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

Edit: changed diagram to name points

edited Jun 12 '20 at 10:38

Community

answered May 03 '16 at 15:22

peterwhy

Not true. http://i.imgur.com/wZuEoN7.jpg – blackcat52 May 03 '16 at 15:32
@blackcat52 I didn't say $\triangle ADC$ (in your image) is isosceles, nor say $\alpha = 20^\circ$. – peterwhy May 03 '16 at 15:34
The ADC in question, the angle of 140 degrees. If the DAC angle is 20, 20 degrees angle of DCA. Therefore, the ADC is isosceles. – blackcat52 May 03 '16 at 15:45
@blackcat52 Again, it is NOT true, and I did NOT imply, that $\alpha = \angle DCA = 20^\circ$. – peterwhy May 03 '16 at 15:52

score 0 · Answer 2 · answered May 03 '16 at 16:06

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 $30^{\circ}$

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

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

2 Answers2