∫∠B = 60

∴∠ AOC = 120 (ipsilateral circumferential angle is equal to half of central angle)

δδAOC

A0 and c0 are two radii of circle O.

∴∠oac=∠oc=( 180- 120)/2 = 30

A vertical line OE passing through point O as a straight line AC intersects with AC at point E.

∫△OAC is an isosceles triangle.

The ∴ median vertical line of AC with OE as the straight line.

That is AE = EC = AC/2 =1.5cm.

In the right triangle AEO

Ae = OA * cos ∠ OAE = OA * cos 30。

Can be pushed

OA=AE/cos30 =√3cm

(√ is the root sign)