The sign shown below consists of two uniform legs attached by a frictionless hinge. The coefficient of friction between the ground and the legs is $\mu$. Which of the following gives the maximum value of $\theta$ such that the sign will not collapse?

$\textbf{(A) } \sin \theta = 2 \mu \\
\textbf{(B) }  \sin \theta /2 = \mu / 2\\
\textbf{(C) } \tan \theta / 2 = \mu\\
\textbf{(D) } \tan \theta = 2 \mu \\
\textbf{(E) } \tan \theta / 2 = 2 \mu$
1. **Identify the forces and torques involved:**
   - The sign consists of two uniform legs attached by a frictionless hinge.
   - The coefficient of friction between the ground and the legs is $\mu$.
   - Let $F$ denote the frictional force.
   - The normal force $F_n$ on each leg is $\frac{Mg}{2}$, where $M$ is the mass of the sign and $g$ is the acceleration due to gravity.

2. **Calculate the torque due to gravity:**
   - Consider the left leg as the center of rotation.
   - The torque due to gravity acting on the center of mass of the left leg is:
     \[
     \tau_{\text{gravity}} = \frac{L}{2} \cdot \frac{Mg}{2} \cdot \sin \frac{\theta}{2}
     \]
     where $L$ is the length of the leg and $\theta$ is the angle between the legs.

3. **Calculate the torque due to friction:**
   - The frictional force on the right leg contributes a torque:
     \[
     \tau_{\text{friction}} = L \cdot F \cdot \cos \frac{\theta}{2}
     \]
   - Since $F_n = \frac{Mg}{2}$, the frictional force $F$ is:
     \[
     F = \mu \cdot F_n = \mu \cdot \frac{Mg}{2}
     \]

4. **Set up the equilibrium condition:**
   - For the sign to be in equilibrium, the torques must balance:
     \[
     \frac{L}{2} \cdot \frac{Mg}{2} \cdot \sin \frac{\theta}{2} = L \cdot \frac{\mu Mg}{2} \cdot \cos \frac{\theta}{2}
     \]

5. **Simplify the equation:**
   - Cancel out common terms:
     \[
     \frac{1}{2} \cdot \frac{Mg}{2} \cdot \sin \frac{\theta}{2} = \frac{\mu Mg}{2} \cdot \cos \frac{\theta}{2}
     \]
     \[
     \frac{1}{4} \cdot \sin \frac{\theta}{2} = \frac{\mu}{2} \cdot \cos \frac{\theta}{2}
     \]

6. **Solve for $\tan \frac{\theta}{2}$:**
   - Divide both sides by $\cos \frac{\theta}{2}$:
     \[
     \frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}} = 2\mu
     \]
     \[
     \tan \frac{\theta}{2} = 2\mu
     \]

7. **Conclusion:**
   - The maximum value of $\theta$ such that the sign will not collapse is given by:
     \[
     \boxed{\textbf{(E) } \tan \theta / 2 = 2 \mu}
     \]