# Find triangle vertices given center and orientation

If you're given the coordinates of the center of a triangle, the distance from the center to each vertex, and the angle above the horizontal and the line connecting the center and one vertex, how do you find the coordinates of each vertex of the triangle?

I want to find the coordinates of A, B, and C:

• The second image suggests we have an isosceles triangle with $AB=AC$ and a point $O$ lying on the median from $A$ to $BC$. Hence, $OB=OC$, but I think $B$ and $C$ could be any points on a circle centred on $O$ with radius $d$, provided $B$ and $C$ are reflections in the line $OA$. (Also, "center of a triangle"... lol.) Apr 11, 2017 at 14:43
• Remembering that any three points form a triangle, unless they are in a straight line, think about a physical implementation of this. You have a fixed point defined for the centre, and then a rod of given distance for each point, which can rotate about the centre. You fix one point with the angle value, but the other two points are still free to move, so they cannot be determined. Apr 11, 2017 at 15:09
• There are many meanings of "center of a triangle." mathsisfun.com/geometry/triangle-centers.html Apr 11, 2017 at 19:40

1. Draw a horizontal vector of length $|OA|$: this vector has coordinates $(|OA|,0)$.
2. Rotate the vector so that it makes angle $\theta$ with the horizontal. Let's call this rotation operation $R_{\theta}$. This gives us the vector $R_\theta (|OA|, 0)$
3. Translate the vector so that it begins at the triangle center $O$: $$O + R_{\theta}(|OA|,0).$$
4. Repeat steps 1-3 for the other corners $B$ and $C$.
Finally recall that counterclockwise rotation by $\theta$ is given by the formula $$R_{\theta}(a,b) = (a\cos\theta - b\sin\theta, a\sin\theta + b\cos \theta)$$ which you can use to simplify your formulas.