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⚡ AZUCATION SHORTCUT:
In this standard configuration (tangent $AB$ between two fixed tangents), the angle $\angle AOB$ is always half of the supplementary angle of $\angle APB$.
Formula: $\angle AOB = \frac{1}{2} \angle QOR$. Since $\angle QOR = 180^\circ - \angle APB$, we get:
$\angle APB = 180^\circ - 2(\angle AOB) = 180^\circ - 100^\circ = \mathbf{80^\circ}$.

Step 1: Understand the Geometry
Let $T$ be the point where the tangent $AB$ touches the circle. From external points, tangents to a circle are equal in length and subtend equal angles at the center.
$\implies \triangle AOQ \cong \triangle AOT$ and $\triangle BOR \cong \triangle BOT$.

Step 2: Relate the Angles at Center
Let $\angle QOA = \angle TOA = \alpha$ and $\angle ROB = \angle TOB = \beta$.
The given angle $\angle AOB = \angle TOA + \angle TOB = \alpha + \beta = 50^\circ$.
The total angle $\angle QOR = 2\alpha + 2\beta = 2(\alpha + \beta) = 2(50^\circ) = 100^\circ$.

Step 3: Solve for $\angle APB$
In quadrilateral $OQPR$, $\angle OQP = \angle ORP = 90^\circ$ (Radius $\perp$ Tangent).
The sum of angles in a quadrilateral is $360^\circ$:
$\angle APB + \angle QOR + 90^\circ + 90^\circ = 360^\circ$
$\angle APB + 100^\circ = 180^\circ$
$\angle APB = \mathbf{80^\circ}$.

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