# Can I prove that a right angle triangle of sides $a$ and $b$ has a angles $θ$ and $Φ$ without trigonometry?

I was looking at the proofs for $$\sin30^{\circ}$$, $$\cos 30^{\circ}$$, $$\tan 30^{\circ}$$, $$\sin 60^{\circ}$$, $$\cos 60^{\circ}$$, and $$\tan 60^{\circ}$$ with the use of this triangle:

The only problem is it assumes that a triangle with lengths $$\frac{1}{2}$$ and $$\frac{\sqrt{3}}{2}$$ has angles of $$30$$ and $$60$$ in those places. It can be proven that those angles are there based on those sides using trigonometry, but then that becomes tautological. So, I will need to prove those angles using some other method.

• Take an equilateral triangle of sidelength $1$ and draw an altitude.
– lulu
Aug 7, 2021 at 17:46
• @lulu: Even simpler might be an equilateral triangle of side length $2$.
– Joe
Aug 7, 2021 at 18:11

Consider following equilateral triangle of side length $$a$$ units
Now $$AB=BC=CA=a$$ and $$BD=0.5a$$, therefore By Pythagoras theorem $$AD=\frac{\sqrt3}{2}a$$
Thus now you get your right triangle $$ABD$$ similar to what you wanted where you know side lengths and angles involved, thus you can prove required trigonometrical results
Just mirror triangle about the side with measure $$\frac{\sqrt 3}2$$, you get an equilateral triangle all its angles equal to $$60^0$$. The altitude on side with measure $$\frac 12$$ bisects the corresponding vertex angle which becomes $$30^o$$