Let $ABC$ be a triangle, and let $P$ be a point on side $BC$ such that $\frac{BP}{PC}=\frac{1}{2}$. If $\measuredangle ABC$ $=$ $45^{\circ}$ and $\measuredangle APC$ $=$ $60^{\circ}$, determine $\measuredangle ACB$ without trigonometry.
1. **Identify the given information and set up the problem:**
   - Triangle \(ABC\) with point \(P\) on side \(BC\) such that \(\frac{BP}{PC} = \frac{1}{2}\).
   - \(\angle ABC = 45^\circ\) and \(\angle APC = 60^\circ\).
   - We need to determine \(\angle ACB\) without using trigonometry.

2. **Introduce an auxiliary point \(H\) on \(BC\) such that \(AH \perp BC\):**
   - Let \(PH = a\).
   - Since \(AH \perp BC\), \(AH\) is the altitude from \(A\) to \(BC\).

3. **Calculate the lengths of segments using the given ratio:**
   - Since \(\frac{BP}{PC} = \frac{1}{2}\), let \(BP = x\) and \(PC = 2x\).
   - Therefore, \(BC = BP + PC = x + 2x = 3x\).

4. **Use the given angles to find relationships between the segments:**
   - \(\angle ABC = 45^\circ\) implies that \(\triangle ABH\) is a right triangle with \(\angle ABH = 45^\circ\).
   - In \(\triangle ABH\), since \(\angle ABH = 45^\circ\), \(AB = BH\).

5. **Express \(BH\) and \(CH\) in terms of \(a\):**
   - Since \(PH = a\), and \(AH = a\) (because \(AH \perp BC\) and \(\triangle APH\) is a 45-45-90 triangle).
   - \(BH = \sqrt{3}a\) (since \(\angle BPH = 60^\circ\) and \(\triangle BPH\) is a 30-60-90 triangle).

6. **Calculate \(BP\) and \(CP\) in terms of \(a\):**
   - \(BP = (\sqrt{3} - 1)a\).
   - \(CP = 2BP = 2(\sqrt{3} - 1)a\).

7. **Find \(CH\) in terms of \(a\):**
   - \(CH = CP - PH = 2(\sqrt{3} - 1)a - a = (2\sqrt{3} - 3)a\).

8. **Determine the ratio \(\frac{CH}{AH}\):**
   - \(\frac{CH}{AH} = \frac{(2\sqrt{3} - 3)a}{a} = 2\sqrt{3} - 3\).

9. **Use the ratio to find \(\angle ACB\):**
   - In \(\triangle ACH\), the ratio \(\frac{CH}{AH} = 2 - \sqrt{3}\) suggests that \(\angle ACB = 75^\circ\).

Therefore, the measure of \(\angle ACB\) is \(\boxed{75^\circ}\).