Critical angle for slope mass descending Steps:

• Component of gravity pulling slope mass m down incline: m g sinθ.
• Tension from hanging mass M equals M g at equilibrium (a=0).
• Set forces equal for limiting case: m g sinθ = M g.
• Solve for angle: sinθ = M/m.

Why C is correct:

• Newton's second law at equilibrium requires parallel component of slope mass weight to balance hanging mass weight, yielding sinθ = M/m as critical condition; above this θ, slope mass accelerates down.

Why the others are wrong:

• A: Reverses mass roles, giving condition for hanging mass descending.
• B: cosθ relates to normal force, irrelevant to along-slope motion.
• D: Inequality involves perpendicular component, not determining descent direction.

Final answer: C