In a circle with centre O, PAX and PBY are the tangents to the circle at points A and B, from an external point P. Q is any point on the circle such that ∠QAX = 59° and ∠QBY = 72°. What is the measure of ∠AQB ?
Answer & explanation
Correct answer: option 3
\(\angle\)OAX = \({90}^\circ\)
= \(\angle\)OAQ = \({90}^\circ\) - \({59}^\circ\)
= \(\angle\)OAQ = \({31}^\circ\)
Now, \(\angle\)OBX = \({90}^\circ\)
= \(\angle\)OBQ = \({90}^\circ\) - \({72}^\circ\)
= \(\angle\)OBQ = \({18}^\circ\)
Since, OB = OQ = OA
= \(\angle\)OBQ = \(\angle\)OBQ = \({18}^\circ\) (radius)
= \(\angle\)OAQ = \(\angle\)OQA = \({31}^\circ\) (radius)
\(\angle\)AQB = \(\angle\)OBQ + \(\angle\)OAQ
= \(\angle\)AQB = (\({31}^\circ\) + \({18}^\circ\))
= \(\angle\)AQB = \({49}^\circ\)
Therefore, \(\angle\)AQB is \({49}^\circ\).