O is the centre of a circle to which PAX and PBY are tangents from point P at points A and B.Q is a point on the circle, such that $$\angle$$QAX = $$49^\circ$$ and $$\angle$$QBY = $$62^\circ$$. What is the measure of $$\angle$$AQB ?

Solution


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