In a 30-60 right triangle the side opposite the 30 degree angle is half the length of the hypotenuse.

A statement from the trigonometry section of Simmons' Precalculus in a nutshell. Please explain.


Here's a friendly equilateral triangle:

The sides are all of the same length - let's say $a$. The angles are all the same too, and since the angles must add up to $180^\circ$, we conclude that the three angles in the equilateral triangle are equal to $180^\circ/3=60^\circ$.

Now we do something sneaky. We draw a line all the way down from the top vertex of the triangle to the midpoint of the bottom line.

This new line cuts our equilateral triangle in half. What are the angles in one half?

  • The angle at the bottom is $90^\circ$.
  • One of the angles is the same as one of the angles in the original equilateral triangle, so it is $60^\circ$.
  • So the third angle must be $180^\circ-90^\circ-60^\circ=30^\circ$.

Now the hypotenuse of this new triangle is $a$, the side length of the equilateral triangle. And the length of the shortest side is $a/2$, since the line we drew cut the bottom line in half.

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    $\begingroup$ Can you also explain how to distinguish friendly equilateral triangles from unfriendly ones? $\endgroup$ – msouth Apr 27 '15 at 5:28
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    $\begingroup$ @msouth: If the triangle growls at you when you approach it, it's probably not friendly and you should not try to pet it. $\endgroup$ – Ilmari Karonen Apr 29 '15 at 9:43
  • $\begingroup$ @msouth, I think he just meant that equilateral triangles are friendly in general because their properties make them easier to deal with in many contexts than another triangles, there different types of equilateral triangles as far as I know. $\endgroup$ – jeremy radcliff May 1 '15 at 1:54
  • $\begingroup$ I’m starting to dislike the proofs that are made with “friendly” equilateral triangle. It makes me think that this would not apply to unfriendly triangles. Is there an “unfriendly” version of this proof? This seems like an important one though because Sin(pi/6) = 1/2 is proved using this. $\endgroup$ – Ravi Sep 19 '20 at 3:28

This statement follows from the theorem:

If $BC$ is the hypotenuse of a right-angled triangle $\triangle ABC$, it follows that the median $AM$ (which corresponds to the hypotenuse) is $AM = \dfrac{BC}{2}$.

Try to apply some basic geometry to the triangles, which are created.

  • $\begingroup$ This is basically Thales' theorem, right? $\endgroup$ – Martin Sleziak Apr 17 '15 at 5:53
  • $\begingroup$ @MartinSleziak I had not Thales' theorem in mind, but the theorem I stated follows from Thales' Theorem. I had another proof in mind for the theorem I mentioned, but the result is all the same. $\endgroup$ – thanasissdr Apr 17 '15 at 8:38

Given the 30,60,90,triangle,rotate the vertex at 90 about the hypotenuse,forming a rectangle whose lateral sides and diagonals form two congruent triangles, both of which are equilateral.Hence,a lateral edge must equal 1/2 of the diagonal.But the diagonal is the hypotenuse. Q.E.D. Edwin Gray

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    $\begingroup$ forming a rectangle whose lateral sides and diagonals form two congruent triangles, both of which are equilateral -- can you draw me a picture of a rectangle made of two equilateral triangles? $\endgroup$ – msouth Apr 27 '15 at 5:26

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