From the figure, it is clear that the angle between OB and AB is $$90^{\circ}$$. That is, OBA forms a right-angled triangle. From trigonometry, we know that $$\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$$ In this case, it becomes $$\cos 60^{\circ} = \frac{\text{OB}}{\text{OA}}$$
$$\text{OB} = \text{OA} \times \cos 60^{\circ} = 10 \times \cos 60^{\circ}$$