C

Step 1: Understand the forces involved.
When a force F is applied at an angle θ to the horizontal to pull a block of mass M on a surface with a coefficient of friction µ, the frictional force (f) opposing the motion is given by:
$$ f = \\mu N $$
where N is the normal force.

Step 2: Determine the normal force.
The normal force N is affected by the vertical component of the applied force. The total vertical forces in the vertical direction can be expressed as:
$$ N = Mg - F \\sin(θ) $$
where Mg is the weight of the block.

Step 3: Substitute for N in the frictional force equation.
Thus,
$$ f = \\mu (Mg - F \\sin(θ)) $$

Step 4: Set up the equation of motion for the horizontal direction:
The horizontal component of the applied force must overcome the frictional force:
$$ F \\cos(θ) = f $$

Step 5: Substitute f in this equation:
$$ F \\cos(θ) = \\mu (Mg - F \\sin(θ)) $$

Step 6: Rearrange the equation to isolate F:
Expanding gives us:
$$ F \\cos(θ) + \\mu F \\sin(θ) = \\mu Mg $$
or
$$ F (\cos(θ) + \\mu \\sin(θ)) = \\mu Mg $$
Thus,
$$ F = \\frac{\\mu Mg}{\cos(θ) + \\mu \\sin(θ)} $$

Step 7: Find the angle for minimum force application.
To minimize F, differentiate F with respect to θ and set the derivative to zero to find the angle θ at which F is minimized:
However, a common solution is to use the angle where the derivative indicates a minimum can be found using:
$$ an(θ) = \\mu $$

Conclusion: The minimum value of force required is:
$$ F_{min} = \\frac{\\mu Mg}{\sqrt{1 + \\mu^2}} $$
and the angle θ is:
$$ θ = an^{-1}(\\mu). $$
Hence, the answer is (C).