A circle is centred at O. Two tangents AP and AQ are drawn from an external point A. If $$\angle$$POQ = 118$$^\circ$$, then $$\angle$$PAQ is equal to:

Given, $$\angle$$POQ = 118$$^\circ$$

AP and AQ are tangents from point A to the circle with centre O

$$=$$>  $$\angle$$OPA = 90$$^\circ$$  and   $$\angle$$OQA = 90$$^\circ$$

From quadrilateral OPAQ,

$$\angle$$POQ + $$\angle$$OQA + $$\angle$$PAQ + $$\angle$$OPA = 360$$^\circ$$

$$=$$>  118$$^\circ$$ + 90$$^\circ$$ + $$\angle$$PAQ + 90$$^\circ$$ = 360$$^\circ$$

$$=$$>  298$$^\circ$$ + $$\angle$$PAQ = 360$$^\circ$$

$$=$$>  $$\angle$$PAQ = 62$$^\circ$$

Hence, the correct answer is Option A