In triangle $ABC$, $AB>AC$. The bisectors of $\angle{B},\angle{C}$ intersect the sides $AC,AB$ at $P,Q$, respectively. Let $I$ be the incenter of $\Delta ABC$. Suppose that $IP=IQ$. How much isthe value of $\angle A$?
1. Given a triangle \(ABC\) with \(AB > AC\), the angle bisectors of \(\angle B\) and \(\angle C\) intersect the sides \(AC\) and \(AB\) at points \(P\) and \(Q\) respectively. Let \(I\) be the incenter of \(\Delta ABC\). We are given that \(IP = IQ\) and need to find the value of \(\angle A\).

2. First, note that \(I\) is the incenter, so it is equidistant from all sides of the triangle. The angle bisectors of \(\angle B\) and \(\angle C\) intersect at \(I\), and \(IP\) and \(IQ\) are the distances from \(I\) to the sides \(AC\) and \(AB\) respectively.

3. Since \(IP = IQ\), the distances from \(I\) to the sides \(AC\) and \(AB\) are equal. This implies that the angles formed by these distances with the sides are also equal. Let \(x\) be the angle \(\angle BPC\) and \(y\) be the angle \(\angle CQB\).

4. By the angle bisector theorem, we have:
   \[
   \angle BPC = \angle A + \frac{\angle B}{2}
   \]
   \[
   \angle CQB = \angle A + \frac{\angle C}{2}
   \]

5. Since \(IP = IQ\), we have:
   \[
   \sin(\angle BPC) = \sin(\angle CQB)
   \]
   This implies:
   \[
   \sin\left(\angle A + \frac{\angle B}{2}\right) = \sin\left(\angle A + \frac{\angle C}{2}\right)
   \]

6. The sine function has the property that \(\sin \theta = \sin (\pi - \theta)\). Therefore, we have two cases:
   \[
   \angle A + \frac{\angle B}{2} = \angle A + \frac{\angle C}{2} \quad \text{or} \quad \angle A + \frac{\angle B}{2} + \angle A + \frac{\angle C}{2} = \pi
   \]

7. Simplifying the first case:
   \[
   \frac{\angle B}{2} = \frac{\angle C}{2} \implies \angle B = \angle C
   \]
   This implies \(AB = AC\), which contradicts the given condition \(AB > AC\).

8. Simplifying the second case:
   \[
   \angle A + \frac{\angle B}{2} + \angle A + \frac{\angle C}{2} = \pi
   \]
   \[
   2\angle A + \frac{\angle B + \angle C}{2} = \pi
   \]
   Since \(\angle B + \angle C = \pi - \angle A\), we have:
   \[
   2\angle A + \frac{\pi - \angle A}{2} = \pi
   \]
   \[
   4\angle A + \pi - \angle A = 2\pi
   \]
   \[
   3\angle A = \pi
   \]
   \[
   \angle A = \frac{\pi}{3} = 60^\circ
   \]

Conclusion:
\[
\boxed{60^\circ}
\]