The tangent at a point A on a circle with center O intersects the diameter PQ of the circle, when extended,at point B. If $\angle BAQ = 105^\circ$, then $\angle APQ$ is equal to:

$75^\circ$

As we know,

\(\angle\)PAQ = 90 [PQ is a diameter]

\(\angle\)BAQ = 105  (Given)

= \(\angle\)BAP + \(\angle\)PAQ = 105

= \(\angle\)BAP = 105 - 90 = 15

As we know,

\(\angle\)BAP = \(\angle\)AQP = 15

If \(\Delta \)APQ

\(\angle\)APQ + \(\angle\)PAQ + \(\angle\)AQP = 180

= \(\angle\)APQ + 90 + 15 = 180

Therefore, \(\angle\)APQ = 180 - 90 - 15 = \({75}^\circ\).