Prove that AS KB given that AOB = 140^.

here's the solution for 2.2.

2.2 Prove that $AS \parallel KB$.

Step 1: Calculate $\angle A\hat{S}B$ and $\angle A\hat{K}B$.
The angle at the centre subtended by arc AB is $\angle A\hat{O}B = 140^\circ$.
The angle at the circumference subtended by the same arc AB is $\angle A\hat{S}B$.

$$\angle A\hat{S}B = \frac{1}{2} \angle A\hat{O}B = \frac{1}{2} \times 140^\circ = 70^\circ$$