The angle subtended at the center of a circle by a chord which is equal in length to the radius of the circle is 60°.

To see why, consider a circle with center O and radius r. Let AB be a chord of the circle such that AB = r. Draw the radii OA and OB. Then, triangle OAB is an isosceles triangle with OA = OB = r. Since the base angles of an isosceles triangle are equal, we have ∠OAB = ∠OBA. Let x = ∠OAB = ∠OBA. Then, the angle subtended at the center of the circle by chord AB is ∠AOB = 2x.

Since the sum of the angles in triangle OAB is 180°, we have x + x + ∠AOB = 180°, or 2x + 2x = 180°. Solving for x, we get x = 45°. Therefore, ∠AOB = 2x = 90°.