Let $ABC$ be an acute triangle with orthocenter $H$. Let $M$, $N$ and $R$ be the midpoints of $AB$, $BC$ an $AH$, respectively. If $A\hat{B}C=70^\large\circ$, compute $M\hat{N}R$.
1. **Identify the nine-point circle**: The nine-point circle of a triangle passes through the midpoints of the sides, the feet of the altitudes, and the midpoints of the segments from the orthocenter to the vertices. In this problem, $M$, $N$, and $R$ lie on the nine-point circle of $\Delta ABC$.

2. **Determine the relevant angles**: Given that $\angle ABC = 70^\circ$, we need to find $\angle MNR$.

3. **Use properties of the nine-point circle**: The nine-point circle intersects side $AC$ at the midpoint $X$ and the foot of the altitude $F$ from $B$ to $AC$. Since $X$, $R$, $M$, and $N$ all lie on the nine-point circle, we have $\angle RNM = \angle RXM$.

4. **Analyze the similarity of triangles**: Consider $\Delta AXM$ and $\Delta ACB$. Since $M$ is the midpoint of $AB$, $\Delta AXM \sim \Delta ACB$ by the Midline Theorem (the line segment joining the midpoints of two sides of a triangle is parallel to the third side and half as long).

5. **Determine the orthocenter relationship**: $R$ is the orthocenter of $\Delta AXM$. Therefore, $\angle RXM = \angle HCB$.

6. **Calculate the angle $\angle HCB$**: Since $\angle ABC = 70^\circ$ and $\Delta ABC$ is an acute triangle, the angles at $H$ and $C$ are complementary to the angles at $A$ and $B$. Specifically, $\angle HCB = 90^\circ - \angle ABC = 90^\circ - 70^\circ = 20^\circ$.

7. **Conclude the angle $\angle MNR$**: Since $\angle RNM = \angle RXM$ and $\angle RXM = 20^\circ$, it follows that $\angle MNR = 20^\circ$.

The final answer is $\boxed{20^\circ}$.