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:

Let $$\angle AQO = x$$

$$ \angle AOP= 2x$$

from the above figure BA is tangent 

then $$\ angle OAB = 90^\circ $$

$$ \angle OAQ = 105^\circ - 90 ^\circ $$

                     = $$15 ^\circ$$  (Given $$\angle BAQ = 105^\circ $$) 

So $$ \angle OQA = \angle OAQ = 15^\circ $$

then $$ \triangle AOP $$, 

$$\angle OPA + \angle OAP + 30^\circ = 180^\circ $$

$$\Rightarrow 2\angle OPA = 180^\circ -30^\circ = 150^\circ $$

$$\Rightarrow \angle OPA = \angle APQ = 75^\circ $$ Ans