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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:

Triangle to prove trig identities

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.

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

2 Answers 2

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Consider following equilateral triangle of side length $a$ units

enter image description here

Note that the brown line segments are medians as well as altitudes as well angle bisectors as well as perpendicular bisectors of the triangle (Since in an equilateral triangle all the 4 major centers of a triangle coincide)

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

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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 $30^o$

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