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Given,
$$\angle$$QAX = $$49^\circ$$
$$\angle$$QBY = $$62^\circ$$
PAX and PBY are tangents at A and B respectively to the circle
$$=$$> $$\angle$$OAX = $$90^\circ$$ and $$\angle$$OBY = $$90^\circ$$
$$\angle$$OAX = $$90^\circ$$
$$=$$> $$\angle$$QAX + $$\angle$$QAO = $$90^\circ$$
$$=$$> $$49^\circ$$ + $$\angle$$QAO = $$90^\circ$$
$$=$$> $$\angle$$QAO = $$41^\circ$$
$$\angle$$OBY = $$90^\circ$$
$$=$$> $$\angle$$QBY + $$\angle$$QBO = $$90^\circ$$
$$=$$> $$62^\circ$$ + $$\angle$$QBO = $$90^\circ$$
$$=$$> $$\angle$$QAO = $$28^\circ$$
Angles opposite to equal sides in a triangle are equal
OA = OQ
$$=$$> $$\angle$$AQO = $$41^\circ$$
OB = OQ
$$=$$> $$\angle$$BQO = $$28^\circ$$
$$\therefore\ $$ $$\angle$$AQB = $$\angle$$AQO + $$\angle$$BQO = $$41^\circ$$ + $$28^\circ$$ = $$69^\circ$$
Hence, the correct answer is Option D