Ajay is standing at point $A$ near Pontianak, Indonesia, $0^\circ$ latitude and $110^\circ \text{ E}$ longitude. Billy is standing at point $B$ near Big Baldy Mountain, Idaho, USA, $45^\circ \text{ N}$ latitude and $115^\circ \text{ W}$ longitude. Assume that Earth is a perfect sphere with center $C$. What is the degree measure of $\angle ACB$?

$
\textbf{(A) }105 \qquad
\textbf{(B) }112\frac{1}{2} \qquad
\textbf{(C) }120 \qquad
\textbf{(D) }135 \qquad
\textbf{(E) }150 \qquad
$
To find the degree measure of $\angle ACB$, we need to convert the spherical coordinates of points $A$ and $B$ into Cartesian coordinates and then use the dot product to find the angle between the vectors.

1. **Convert Point $A$ to Cartesian Coordinates:**
   - Point $A$ is at $(1, 90^\circ, 0^\circ)$ in spherical coordinates.
   - The Cartesian coordinates $(x, y, z)$ can be found using:
     \[
     x = \rho \sin \theta \cos \phi
     \]
     \[
     y = \rho \sin \theta \sin \phi
     \]
     \[
     z = \rho \cos \theta
     \]
   - For $A$:
     \[
     x_A = 1 \cdot \sin 90^\circ \cdot \cos 0^\circ = 1 \cdot 1 \cdot 1 = 1
     \]
     \[
     y_A = 1 \cdot \sin 90^\circ \cdot \sin 0^\circ = 1 \cdot 1 \cdot 0 = 0
     \]
     \[
     z_A = 1 \cdot \cos 90^\circ = 1 \cdot 0 = 0
     \]
   - Therefore, $\vec{A} = (1, 0, 0)$.

2. **Convert Point $B$ to Cartesian Coordinates:**
   - Point $B$ is at $(1, 45^\circ, 135^\circ)$ in spherical coordinates.
   - For $B$:
     \[
     x_B = 1 \cdot \sin 45^\circ \cdot \cos 135^\circ = 1 \cdot \frac{\sqrt{2}}{2} \cdot \left(-\frac{\sqrt{2}}{2}\right) = -\frac{1}{2}
     \]
     \[
     y_B = 1 \cdot \sin 45^\circ \cdot \sin 135^\circ = 1 \cdot \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} = \frac{1}{2}
     \]
     \[
     z_B = 1 \cdot \cos 45^\circ = 1 \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2}
     \]
   - Therefore, $\vec{B} = \left(-\frac{1}{2}, \frac{1}{2}, \frac{\sqrt{2}}{2}\right)$.

3. **Calculate the Dot Product $\vec{A} \cdot \vec{B}$:**
   \[
   \vec{A} \cdot \vec{B} = (1, 0, 0) \cdot \left(-\frac{1}{2}, \frac{1}{2}, \frac{\sqrt{2}}{2}\right) = 1 \cdot \left(-\frac{1}{2}\right) + 0 \cdot \frac{1}{2} + 0 \cdot \frac{\sqrt{2}}{2} = -\frac{1}{2}
   \]

4. **Calculate the Magnitudes of $\vec{A}$ and $\vec{B}$:**
   \[
   |\vec{A}| = \sqrt{1^2 + 0^2 + 0^2} = 1
   \]
   \[
   |\vec{B}| = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{1}{4} + \frac{2}{4}} = \sqrt{1} = 1
   \]

5. **Use the Dot Product to Find the Angle $\theta$:**
   \[
   \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta
   \]
   \[
   -\frac{1}{2} = 1 \cdot 1 \cdot \cos \theta
   \]
   \[
   \cos \theta = -\frac{1}{2}
   \]
   \[
   \theta = \cos^{-1}\left(-\frac{1}{2}\right) = 120^\circ
   \]

The final answer is $\boxed{120^\circ}$.