**most important**screen of the module. So, be sure that you understand everything in it!

If we know the

**magnitude**and the

**angle**, calculating the x-component is simple

**trigonometry**.

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 \(\cos 60^{\circ} = \frac{\text{OB}}{\text{OA}}\)

Thus, the x-component, which is nothing but the length OB, is

\(\text{OB} = \text{OA} \times \cos 60^{\circ} = 10 \times \cos 60^{\circ}\)