Two tangents drawn from a point $P$ touch a circle with center $O$ at points $Q$ and $R$. Points $A$ and $B$ lie on $PQ$ and $PR$, respectively, such that $AB$ is also a tangent to the same circle. If $\angle AOB = 50^\circ$, then $\angle APB$, in degrees, equals

Two tangents $PQ$ and $PR$ are drawn from an external point $P$ to a circle with center $O$, touching the circle at $Q$ and $R$. Points $A$ and $B$ lie on $PQ$ and $PR$ respectively, and $AB$ is also tangent to the circle. Let $AB$ touch the circle at point $T$.

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Since two tangents drawn from an external point to a circle are equal in length, the line joining that external point to the center bisects the angle between the two tangents.

From point $A$, the two tangents to the circle are $AQ$ and $AT$.

So $OA$ bisects $\angle QAT$, meaning $\angle OAQ = \angle OAT = \alpha$.

From point $B$, the two tangents to the circle are $BR$ and $BT$.

So $OB$ bisects $\angle RBT$, meaning $\angle OBR = \angle OBT = \beta$.

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Since $T$ lies on segment $AB$, we get $\angle QAB = \angle QAT = 2\alpha$.

Since $A$ lies on line $PQ$:

$\angle PAB = 180^\circ - 2\alpha$

Similarly:

$\angle PBA = 180^\circ - 2\beta$

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In $\triangle OAB$:

$\angle OAB + \angle OBA + \angle AOB = 180^\circ$

$\alpha + \beta = 180^\circ - 50^\circ = 130^\circ$

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In $\triangle PAB$:

$\angle APB + \angle PAB + \angle PBA = 180^\circ$

$\angle APB + (180^\circ - 2\alpha) + (180^\circ - 2\beta) = 180^\circ$

$\angle APB = 2(\alpha + \beta) - 180^\circ$

$\angle APB = 2(130^\circ) - 180^\circ$

$\angle APB = 80^\circ$