If PA and PB are tangents drawn from an external point P to a circle with centre O such that $$\angle$$APB = 70$$^\circ$$, then         $$\angle$$OAB is equal to:

Given, $$\angle$$APB = 70$$^\circ$$

PA and PB are the tangents to the circle with centre O

$$\Rightarrow$$ $$\angle$$OAP = 90$$^\circ$$ and $$\angle$$OBP = 90$$^\circ$$

In quadrilateral OAPB,

$$\angle$$AOB + $$\angle$$OBP + $$\angle$$APB + $$\angle$$OAP = 360$$^\circ$$

$$\Rightarrow$$ $$\angle$$AOB + 90$$^\circ$$ + 70$$^\circ$$ + 90$$^\circ$$ = 360$$^\circ$$

$$\Rightarrow$$ $$\angle$$AOB + 250$$^\circ$$ = 360$$^\circ$$

$$\Rightarrow$$ $$\angle$$AOB = 110$$^\circ$$

In $$\triangle\ $$OAB, OA = OB

Angles opposite to equal sides are equal in triangle

$$\Rightarrow$$ $$\angle$$OBA = $$\angle$$OAB

In $$\triangle\ $$OAB,

$$\angle$$AOB + $$\angle$$OBA + $$\angle$$OAB = 180$$^\circ$$

$$\Rightarrow$$ 110$$^\circ$$ + $$\angle$$OAB + $$\angle$$OAB = 180$$^\circ$$

$$\Rightarrow$$ 2$$\angle$$OAB = 70$$^\circ$$

$$\Rightarrow$$ $$\angle$$OAB = 35$$^\circ$$

Hence, the correct answer is Option A