1. **Problem statement:** From an external point $P$, tangents $PA$ and $PB$ are drawn to a circle with center $O$. Given that $\angle PAB = 50^\circ$, find $\angle AOB$. 2. **Key properties and formulas:** - Tangents from an external point to a circle are equal in length. - The radius drawn to the point of tangency is perpendicular to the tangent line. - $\triangle PAB$ is formed by the tangents and chord $AB$. - $\angle AOB$ is the central angle subtending chord $AB$. 3. **Step-by-step solution:** - Since $PA$ and $PB$ are tangents from $P$, $PA = PB$. - $\triangle PAB$ is isosceles with $PA = PB$. - Given $\angle PAB = 50^\circ$, then $\angle PBA = 50^\circ$ (base angles of isosceles triangle). - Sum of angles in $\triangle PAB$: $$\angle PAB + \angle PBA + \angle APB = 180^\circ$$ $$50^\circ + 50^\circ + \angle APB = 180^\circ$$ $$\angle APB = 180^\circ - 100^\circ = 80^\circ$$ - Since $OA$ and $OB$ are radii, $\triangle OAB$ is isosceles with $OA = OB$. - The angle between the tangents at $P$ is $\angle APB = 80^\circ$. - The quadrilateral $OAPB$ has $OA \perp PA$ and $OB \perp PB$ because radius is perpendicular to tangent. - Therefore, $\angle OAP = 90^\circ$ and $\angle OBP = 90^\circ$. - In $\triangle APB$, angles at $A$ and $B$ are $50^\circ$ each, and $\angle APB = 80^\circ$. - The central angle $\angle AOB$ subtends the same chord $AB$ as the angle $\angle APB$ at the circumference but on the opposite side. - The central angle is twice the angle at the circumference subtending the same chord: $$\angle AOB = 2 \times \angle APB = 2 \times 80^\circ = 160^\circ$$ 4. **Final answer:** $$\boxed{\angle AOB = 160^\circ}$$