Can there be two triangles, for example $ \triangle ABC $ and $ \triangle DEF$ so that the measure of their sides is the same but their angles are different? Thanks.

  • 3
    $\begingroup$ See this. $\endgroup$ Sep 25, 2017 at 11:53
  • 7
    $\begingroup$ Would a mirrored triangle count as the same or different? $\endgroup$
    – Pieter B
    Sep 25, 2017 at 14:15
  • 2
    $\begingroup$ The condition that 2 triangles are completely identical, is governed by the law of congreuncy of triangles. If 3 sides are equal then the triangles are congruent and therefore equal in every aspect. $\endgroup$ Sep 26, 2017 at 6:28

4 Answers 4


Yes, if the triangles aren't necessarily both Euclidean. This may not be what you meant, and the other answers may answer your question more appropriately, but this seems a good opportunity to point out some more exotic scenarios that can exist.

One thing we could do is take triangles from different geometries. For instance, the sum of the angles of a triangle drawn on a sphere will always be greater than $180$ degrees. So you could draw one triangle on a flat plane, another on a sphere, make the side lengths match, and you will get different sized angles.

Be aware though that it isn't sufficient to just scale the metric on a Euclidean space. That is, if I define two different planes where distance is greater in one but they are both flat, and I draw two triangles with matching side length, the angles will still be the same. The important thing about the example of the previous paragraph is that we changed the curvature of the space.

Maybe it's cheating to allow the two triangles to come from different geometries. Well it is in fact possible to draw two triangles with matching side lengths and different angles, in the same geometry. You can do this in the hyperbolic plane by allowing vertices on the boundary of the plane.

The boundary of the hyperbolic plane is like a circle that is infinitely far away from any point in the space. If you put two vertices on this boundary, then pick whatever point you want within the space for your third vertex, each side of the triangle will have infinite length. So the lengths of the sides would be unaffected by where you choose to put your third vertex. Also, the two angles at the vertices on the boundary would actually be $0$ degrees, and the third one could be any number up to and including $180$ degrees, depending on where you put it! This wouldn't work in Euclidean space. You can also talk about a circle boundary of $\mathbb{R}^2$ infinitely far away, but there, you can't draw a straight line connecting arbitrary boundary points and passing through the plane.

There's lots of Google-able info out there (and in Wikipedia too) about the hyperbolic plane, if this concept interests you and you want to know more!

  • 2
    $\begingroup$ Interesting, but the main-case answer is still no. Maybe it is helpful to give some concrete numbers. To your first example, consider a triangle with side lengths $a=1$, $b=1$, and $c=1$. In the Euclidean plane, all angles will be $60^\circ$. On the sphere with curvature (positive) one, all angles will be $69.47^\circ$. The angular excess corresponds to the "total" positive curvature within the triangle. For your second example, you are saying that sides $a=1$, $b=\infty$, and $c=\infty$ may lead to different (non-congruent) triangles in the hyperbolic plane extended with ideal points. $\endgroup$ Sep 25, 2017 at 8:23
  • 5
    $\begingroup$ A silly example is to consider a general regular surface in $\mathbb{R}^3$, or just a general two-dimensional Riemannian manifold. Here the space can be different in different "regions". In particular, the Gauss curvature can vary, so it is obvious that a triangle in one place with sides $a$, $b$ and $c$ will not have the same angles as a triangle in another place in the surface with the same sides $a$, $b$ and $c$. $\endgroup$ Sep 25, 2017 at 8:26
  • 1
    $\begingroup$ @JeppeStigNielsen No, $a=b=c=\infty$ because each edge touches at least one ideal point. Of course the "main-case" answer is no but I'm hoping to pique the OP's curiosity about more interesting stuff. $\endgroup$
    – j0equ1nn
    Sep 25, 2017 at 9:17
  • 1
    $\begingroup$ nit: Is it really a Euclidean space if you replace the Euclidean norm with a different metric? What would the idea of a flat manifold be if you decouple it from the associated norm? $\endgroup$
    – Thoth19
    Sep 25, 2017 at 21:40
  • 1
    $\begingroup$ @Thoth19 I've made a slight edit to improve that part, thanks! $\endgroup$
    – j0equ1nn
    Sep 25, 2017 at 22:40

No: given three sides of a triangle, the cosine rule determines the angles unambiguously: $$ c^2 = a^2+b^2 - 2ab\cos{\gamma}, $$ so $$ \cos{\gamma} = \frac{a^2+b^2-c^2}{2ab}, $$ where $\gamma$ is one of the angles, $c$ is the side opposite $\gamma$, and $a,b$ the other two sides. Cosine is one-to-one from the angles between $0$ and $\pi$ to the real interval $(-1,1)$, and one can check that the right-hand side lies strictly between $-1$ and $1$ when $a,b,c$ form a triangle, so one has $$ \gamma = \arccos{\left( \frac{a^2+b^2-c^2}{2ab} \right)}. $$

  • $\begingroup$ I think one doesn't have to check that the right-hand side lies in $(-1, 1)$, because we know it is equal to $\cos \gamma$. The point that needs to be made here instead is that $\gamma \in (0, \pi)$, for that allows us to pass from $\cos \gamma = r$ to $\gamma = \arccos r$. $\endgroup$
    – Adayah
    Sep 26, 2017 at 10:18
  • $\begingroup$ Given three numbers satisfying the triangle inequality, it's not immediately obvious that "$\cos{\gamma}$" is the cosine of an actual angle (i.e. that they actually are the sides of a triangle), so that does need proof (equally in the spherical and hyperbolic cases, with their own cosine rules). $\endgroup$
    – Chappers
    Sep 26, 2017 at 13:43
  • $\begingroup$ Maybe I understand you incorrectly, but: in the beginning we get two triangles, both with sides $a, b, c$, and we're supposed to show that their angles opposite to the side of length $c$ are equal. So it is immediately obiovus that $a, b, c$ form a triangle. Anyway, I don't see why we would need that: we just say that $\cos \gamma_1 = \frac{a^2+b^2-c^2}{2ab} = \cos \gamma_2$ hence $\gamma_1 = \gamma_2$ because both angles lie in $(0, \pi)$ where $\cos$ is 1-1. $\endgroup$
    – Adayah
    Sep 26, 2017 at 19:17
  • $\begingroup$ A similar law of cosines would show this in spherical or hyperbolic geometry. By comparison, the counterexamples in my answer rely on either varying curvature from one triangle to the other, or using ideal vertices. $\endgroup$
    – j0equ1nn
    Sep 26, 2017 at 22:39
  • $\begingroup$ I suppose this was the answer I was needing. I also wanted to know why isosceles triangles have a pair of equal angles. Using the cosine law, I think it´s possible to do it. Here´s what I did: $\endgroup$
    – Vmimi
    Oct 2, 2017 at 23:13

No, this is not possible in normal Euclidean geometry. Two triangles are congruent if all the corresponding sides are pairwise equal, so their angles will be equal too.

  • 1
    $\begingroup$ The question did not specifically say that it was considering only Euclidean geometry, so here's a complete summary: "no" in elliptic/spherical/Euclidean geometry, but "yes" in hyperbolic geometry. $\endgroup$ Sep 25, 2017 at 5:37
  • 4
    $\begingroup$ @J.M.isnotamathematician Thank you for your useful addition. I answered the question that I thought was being asked, in the most concise fashion I could muster. But your comment is a good completion.:) $\endgroup$
    – Deepak
    Sep 25, 2017 at 6:20
  • 2
    $\begingroup$ @J.M.isnotamathematician I would be careful with that, because the answer is still "no" in hyperbolic geometry unless you allow for ideal vertices, as detailed in my answer below. $\endgroup$
    – j0equ1nn
    Sep 25, 2017 at 6:57
  • $\begingroup$ @j0e, the question was only asking if it was possible. ;) You gave the exact conditions for it to be possible in hyperbolic space, so your answer is a good elaboration. $\endgroup$ Sep 25, 2017 at 7:36
  • 1
    $\begingroup$ @J.M.isnotamathematician Yeah but I'm saying, what I did is especially weird, using ideal vertices. I think it's misleading to just say the SSS property fails in hyperbolic space. Especially if somebody's looking for a quick take-away! $\endgroup$
    – j0equ1nn
    Sep 25, 2017 at 7:42

Take 3 sticks, each must be shorter than the sum of the other 2 sticks. Try to construct more than one triangle, can you? No. If you take 2 sticks and joins their ends, then the third stick will be the constraint for the other ends of the 2 sticks.

  • $\begingroup$ This answer is evidently rather unloved, but I think it is useful to observe that one can see it very intuitively. An alternative but similar approach would be the construction with ruler and compass: this will if feasible yield two positions for the third point, but with the same angles in the reverse direction of rotation. $\endgroup$
    – PJTraill
    Sep 27, 2017 at 21:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .