For an olympiad geometry problem, Tina wants to draw an acute triangle whose angles each measure a multiple of $10^{\circ}$. She doesn't want her triangle to have any special properties, so none of the angles can measure $30^{\circ}$ or $60^{\circ}$, and the triangle should definitely not be isosceles.

How many different triangles can Tina draw? (Similar triangles are considered the same.)

[i]Proposed by Evan Chen[/i]
1. **Identify the possible angle measures:**
   Since each angle must be a multiple of \(10^\circ\) and the triangle must be acute, the possible angles are:
   \[
   10^\circ, 20^\circ, 30^\circ, 40^\circ, 50^\circ, 60^\circ, 70^\circ, 80^\circ
   \]
   However, the problem states that none of the angles can measure \(30^\circ\) or \(60^\circ\). Therefore, the possible angles are:
   \[
   10^\circ, 20^\circ, 40^\circ, 50^\circ, 70^\circ, 80^\circ
   \]

2. **Check for acute triangle conditions:**
   For a triangle to be acute, all its angles must be less than \(90^\circ\). The angles listed above satisfy this condition.

3. **Sum of angles in a triangle:**
   The sum of the angles in any triangle must be \(180^\circ\). We need to find combinations of three distinct angles from the set \(\{10^\circ, 20^\circ, 40^\circ, 50^\circ, 70^\circ, 80^\circ\}\) that add up to \(180^\circ\).

4. **Eliminate invalid angles:**
   - If we include \(10^\circ\) in the triangle, the remaining two angles must sum to \(170^\circ\), which is impossible since the maximum sum of two angles from the set is \(80^\circ + 70^\circ = 150^\circ\).
   - If we include \(20^\circ\) in the triangle, the remaining two angles must sum to \(160^\circ\), which is also impossible for the same reason.

5. **Check remaining combinations:**
   - For \(40^\circ\), the remaining two angles must sum to \(140^\circ\). Possible pairs are \((50^\circ, 90^\circ)\), \((70^\circ, 70^\circ)\), and \((80^\circ, 60^\circ)\). None of these pairs are valid since \(90^\circ\) and \(60^\circ\) are not in the set, and \(70^\circ\) cannot be repeated.
   - For \(50^\circ\), the remaining two angles must sum to \(130^\circ\). Possible pairs are \((70^\circ, 60^\circ)\) and \((80^\circ, 50^\circ)\). Neither pair is valid since \(60^\circ\) is not in the set, and \(50^\circ\) cannot be repeated.
   - For \(70^\circ\), the remaining two angles must sum to \(110^\circ\). Possible pairs are \((40^\circ, 70^\circ)\) and \((50^\circ, 60^\circ)\). Neither pair is valid since \(60^\circ\) is not in the set, and \(70^\circ\) cannot be repeated.
   - For \(80^\circ\), the remaining two angles must sum to \(100^\circ\). Possible pairs are \((20^\circ, 80^\circ)\) and \((40^\circ, 60^\circ)\). Neither pair is valid since \(60^\circ\) is not in the set, and \(80^\circ\) cannot be repeated.

6. **Conclusion:**
   After testing all possible combinations, we find that no set of three distinct angles from \(\{10^\circ, 20^\circ, 40^\circ, 50^\circ, 70^\circ, 80^\circ\}\) sums to \(180^\circ\). Therefore, it is impossible to form such a triangle.

\(\boxed{0}\)