We can draw the following diagram as per the given question :
Now, taking $\triangle AOT$ and $\triangle AOQ$
AO is the common side
AT=AQ (Tangents drawn from an external point are equal in length)
OT=OQ (radius of circle)
So, by S.S.S., $\triangle\ AOT$ is congruent to $\triangle\ AOQ$
So, by C.P.C.T., $\angle\ AOT=\angle\ AOQ$
Similarly, for $\triangle BOT$ and $\triangle BOR$, we can do the same thing.
So, we can say $\triangle BOT$ and $\triangle BOR$ will also be congruent.
Now, $\angle\ QOR=\angle\ AOQ+\angle\ AOB+\angle\ ROB=\angle\ AOT+\angle\ AOB+\angle\ BOT=2\angle\ AOB=2\times\ 50^{\circ\ }=100^{\circ\ }$
Now, in quadrilateral OQPR,
$\angle\ OQP=\angle\ ORP=90^{\circ\ }$ (Tangents drawn to a point is perpendicular to the radius drawn at the same point)
Now, sum of angles of a quadrilateral is $360^{\circ\ }$
So, $\angle\ APB+\angle\ QOR=360^{\circ\ }-\left(90^{\circ\ }+90^{\circ\ }\right)=180^{\circ\ }$
So, $\angle\ APB=180^{\circ\ }-100^{\circ\ }=80^{\circ\ }$
