# Generate two valid vertices of isosceles triangle, given one vertex, an angle, and a distance

Trigonometry question:

I want to find a way to randomly sample the coordinates of the two remaining vertices $$C_2$$ and $$C_3$$ of an isosceles triangle, given one initial coordinate $$C_1$$. I have the coordinate of one vertex ($$C_1$$), the angle between the vertex and the two remaining vertices (let's call it $$\theta$$), and the distance between $$C_1$$ and the two remaining vertices $$C_2$$ and $$C_3$$ (let's call it $$R$$: by isosceles triangle I mean, $$R$$ describes both the distance from $$C_1$$ to $$C_2$$ as well as from $$C_1$$ to $$C_3$$)

How do I randomly choose two valid points $$C_2$$ and $$C_3$$ that satisfy the conditions of being theta degrees apart (relative to $$C_1$$) and being the same distance $$R$$ from $$C_1$$? • Welcome to Math.SE. For future reference, you may take a look at this page to see how to format math on this site. – g.kov Jul 8 '20 at 21:19
• thank you, and will do! – Conor Jul 8 '20 at 21:35

## 1 Answer

Given $$C_1$$, $$\theta$$ and $$|C_1C_2|=|C_1C_3|=R$$, the locus of the points $$C_2,C_3$$ is the circumference of the circle centered at $$C_1$$, and the location of one of the points $$C_2,C_2$$ is defined by the location of the other. So, for example, you can choose randomly an angle $$\phi$$ to locate the point $$C_3$$,

\begin{align} C_3&=C_1+R\cdot(\cos\phi,\sin\phi) \end{align}
then the coordinates of the other point would be \begin{align} C_2&=C_1+R\cdot(\cos(\phi+\theta),\sin(\phi+\theta)) . \end{align}