Let say that I have an equilateral triangle with vertices: (0,0) (200,346.4) (400,0). Say I want to rotate this triangle 30 degrees clockwise, how would I find the new vertices?

Edit1: Rotating about the center of the triangle, my apologies.


closed as off-topic by Chappers, user223391, Harish Chandra Rajpoot, graydad, J. W. Perry Sep 26 '15 at 2:29

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  • $\begingroup$ You should mention the vertex/point or the axis about which the triangle is rotated. $\endgroup$ – Harish Chandra Rajpoot Sep 25 '15 at 20:04
  • $\begingroup$ Are you rotating about the origin? $\endgroup$ – David Quinn Sep 25 '15 at 20:04
  • $\begingroup$ Yes rotating about the center of the triangle, my apologies $\endgroup$ – samuelk71 Sep 25 '15 at 20:04
  • $\begingroup$ Are you looking for an analytic solution or a geometric one? $\endgroup$ – John Douma Sep 25 '15 at 20:06
  • $\begingroup$ The actual vertices I am working on a website. -Thanks $\endgroup$ – samuelk71 Sep 25 '15 at 20:08

First subtract the coordinates of the centre, $(200,\frac{200\sqrt{3}}{3})$ from each vertex coordinate. Then multiply each new vertex by the rotation matrix for clockwise rotation by 30 degrees, i.e. $$\left(\begin{matrix}\cos(-30)&-\sin(-30)\\\sin(-30)&\cos(-30)\end{matrix}\right)$$ to obtain three new rotated vertices. Then add back the centre coordinates to each rotated vertex.

  • $\begingroup$ Nice thats what I was looking for I was actually almost there haha. $\endgroup$ – samuelk71 Sep 25 '15 at 20:34

Here is how I would go about solving this;

  • Imagine the three points of the triangle as points on a circle
    "concentric" with the triangle.
  • These points form three 120 degree arcs.
  • The midpoints of these arcs are the vertices after a 60 degree rotation, and the quarterpoints of these arcs are the possible vertices after the two possible 30 degree rotations.

I am not at a point where I can do the calculations though, sorry. I would do this as a comment if I had that permission available to me on this stack. Hopefully this is able to point you in the right direction though!

  • $\begingroup$ I was thinking something about that, but my roommate suggested something along the lines of a rotational matrix where you would use each line as a vector and rotate them individually. I'll look into both. $\endgroup$ – samuelk71 Sep 25 '15 at 20:16

Rotate the triangle about [b]what[/b] point?

I suspect you mean about the center of the triangle. If that is true, the first thing you need to do is determine that center. Fortunately, for a triangle, that is just the mean of the coordinates of the point, ((0+ 200+ 400)/3, (0+ 346.4/3))= (200, 115.5).

To determine what happens to the coordinates of a vertex when you wrote the triangle about that point: 1) Translate the whole triangle so that the center is moved to the orgin: Subtract the coordinates of the center from the coordinates of the vertex. (0, 0) becomes (-200, -115.5)

2) Rotate around the origin by angle [math]\theta[/tex]: The new x coordinate becomes [math]x'= x cos(\theta)- y sin(\theta)[/itex] and the new y coordinate [math]y'= x sin(\theta)+ y cos(\theta)[/math]. With the values above and [math]\theta= 30[/math] degrees, [math]x'= (-200(0.8660)+ 115.5(.5)= -115.45[/math] and [math]y'= (-2oo)(.5)- 115.5(0.8660)= -200.023[/math]. That gives (-115.45, -200.023)

3) Translate back to the center- add the coordinates of the center: (-115.45+ 200, -220.023+ 115.5)= (84.55, -104.523).


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