Triangle $ PAB$ is formed by three tangents to circle $ O$ and $ < APB \equal{} 40^{\circ}$; then angle $ AOB$ equals:

$ \textbf{(A)}\ 45^{\circ} \qquad\textbf{(B)}\ 50^{\circ} \qquad\textbf{(C)}\ 55^{\circ} \qquad\textbf{(D)}\ 60^{\circ} \qquad\textbf{(E)}\ 70^{\circ}$
1. Given that triangle $PAB$ is formed by three tangents to circle $O$, and $\angle APB = 40^\circ$.
2. Since $PAB$ is a triangle formed by tangents, the points $A$ and $B$ are points of tangency. Therefore, $PA$ and $PB$ are tangents to the circle at points $A$ and $B$ respectively.
3. The tangents to a circle from an external point are equal in length. Thus, $PA = PB$.
4. The angles formed by the tangents and the radii at the points of tangency are right angles. Therefore, $\angle OAP = \angle OBP = 90^\circ$.
5. Since $PA = PB$, triangle $PAB$ is isosceles with $\angle PAB = \angle PBA$.
6. Let $\angle PAB = \angle PBA = x$. Then, the sum of the angles in triangle $PAB$ is:
   \[
   x + x + 40^\circ = 180^\circ
   \]
   Simplifying, we get:
   \[
   2x + 40^\circ = 180^\circ \implies 2x = 140^\circ \implies x = 70^\circ
   \]
   Therefore, $\angle PAB = \angle PBA = 70^\circ$.
7. Now, consider the quadrilateral $AOBP$. Since $PA$ and $PB$ are tangents, $\angle OAP = \angle OBP = 90^\circ$.
8. The sum of the angles in quadrilateral $AOBP$ is $360^\circ$. Therefore:
   \[
   \angle OAP + \angle OBP + \angle APB + \angle AOB = 360^\circ
   \]
   Substituting the known values:
   \[
   90^\circ + 90^\circ + 40^\circ + \angle AOB = 360^\circ
   \]
   Simplifying, we get:
   \[
   220^\circ + \angle AOB = 360^\circ \implies \angle AOB = 140^\circ
   \]
9. However, we need to find the angle $\angle AOB$ in the context of the circle. Since $A$ and $B$ are points of tangency, $\angle AOB$ is the central angle subtended by the arc $AB$.
10. The central angle $\angle AOB$ is twice the inscribed angle $\angle APB$. Therefore:
    \[
    \angle AOB = 2 \times \angle APB = 2 \times 40^\circ = 80^\circ
    \]

The final answer is $\boxed{80^\circ}$