Question:

1. Let $O$ be the centre of the circle, $P$ the external point, and $A$, $B$ the points of contact of the two tangents $PA$ and $PB$.
2. Join $OA$ and $OB$. By the tangent‑radius theorem, $PA \perp OA$ and $PB \perp OB$; therefore $\angle OAP = \angle OBP = 90^{\circ}$.
3. Consider triangle $OAP$. The interior angles satisfy
$$\angle AOP + \angle OAP + \angle OPA = 180^{\circ}.$$
Substituting $\angle OAP = 90^{\circ}$ gives
$$\angle AOP = 180^{\circ} - 90^{\circ} - \angle OPA = 90^{\circ} - \angle OPA.$$
4. Similarly, for triangle $OBP$ we obtain
$$\angle BOP = 90^{\circ} - \angle OPB.$$
5. The angle between the two tangents at $P$ is $\angle APB = \angle OPA + \angle OPB$ (exterior angles of the two right‑angled triangles).
6. Add the two expressions obtained in steps 3 and 4:
\begin{align*}
\angle AOP + \angle BOP &= (90^{\circ} - \angle OPA) + (90^{\circ} - \angle OPB) \\
&= 180^{\circ} - (\angle OPA + \angle OPB) \\
&= 180^{\circ} - \angle APB.
\end{align*}
7. Hence
$$\angle APB + (\angle AOP + \angle BOP) = 180^{\circ}.$$
The angle $\angle AOP + \angle BOP$ is the angle subtended at the centre by the chord $AB$ (i.e., the line‑segment joining the points of contact). Therefore, the angle between the two tangents is supplementary to the central angle subtended by $AB$.
8. Concluding statement: \[\boxed{\angle \text{(tangents)} + \angle \text{(central angle)} = 180^{\circ}}\]