Proving the circumradius (R) for an equilateral triangle

Question

I am intrigued about how people would go about proving the formula for the circumradius in the image above. I went about it by using the cosine rule and it worked out nicely. Any other methods?

Answer 1

You have a right triangle with angles $30,60$ and the hypothenuse is $R$. You can get $R$ from there.

Answer 2

Use Pitagoras Theorem $$R=\sqrt{\left(\frac{s}{2}\right)^2 + \left(\frac{s}{2 \sqrt{3}}\right)^2}=s\frac{\sqrt3}{3}$$