# How to project a 2D shape (Triangle) at a given angle

I am trying to take an equilateral triangle and get it's projection at an angle of 60 degrees. I have the triangle already (SSSTriangle[6, 6, 6]) and I have tried to figure out how to use Projection[] but I have trouble figuring out how to employ this with vectors. I can also get the Radon transform at 60 degrees, but this is still not the actual 2D projection that I am trying to create. Neither of these options seems to be what I really want. How do I create this projection and visualize it? I know drawing it out, it should basically look like another triangle.

Your problem was not well defined, as in 3D there are many ways to rotate object and many ways to project vectors and objects. As a demonstration I rotated you triangle around $$x$$, $$y$$ and $$z$$ axis by 60 degrees clockwise. If you run the code, you can rotate the 3D Graphics and take a look at it under different angles. Then I projected the triangles to the $$xy$$ plane. The code is written such that you can easily change the rotation axis and projection plane.

triangle = SSSTriangle[6, 6, 6]
basisX = {1, 0, 0};
basisY = {0, 1, 0};
basisZ = {0, 0, 1};
triangles3D =
Table[triangle /. {x_, y_} :>
RotationMatrix[60 Degree, v].{x, y, 0},
{v, {basisX, basisY, basisZ}}]
Graphics3D[{FaceForm[None], EdgeForm[Black], triangles3D},
Axes -> True, AxesLabel -> {x, y, z}, ViewPoint -> Top]
triangles2D =
triangles3D /.
p : {x_?NumberQ, y_, z_} :> {
Projection[p, basisX].basisX,
Projection[p, basisY].basisY
}
Graphics[{FaceForm[None], EdgeForm[Black], triangles2D}, Axes -> True,
AxesLabel -> {x, y}]

Triangle[{{0, 0}, {6, 0}, {3, 3 Sqrt[3]}}]

{Triangle[{{0, 0, 0}, {6, 0, 0}, {3, (3 Sqrt[3])/2, 9/2}}],
Triangle[{{0, 0, 0}, {3, 0, -3 Sqrt[3]}, {3/2,
3 Sqrt[3], -((3 Sqrt[3])/2)}}],
Triangle[{{0, 0, 0}, {3, 3 Sqrt[3], 0}, {-3, 3 Sqrt[3], 0}}]}


{Triangle[{{0, 0}, {6, 0}, {3, (3 Sqrt[3])/2}}],
Triangle[{{0, 0}, {3, 0}, {3/2, 3 Sqrt[3]}}],
Triangle[{{0, 0}, {3, 3 Sqrt[3]}, {3, 3 Sqrt[3]}}]}