The value of angle such that the acceleration of A is g/6 downwards along the incline plane.

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Block A (2m) on the 30° face and B (m) on the $θ$ face, connected over a pulley. A moves down with $a = g /6$ , so B moves up its face.

A: $2 m g sin 3 0^{\circ} - T = 2 m \cdot \frac{g}{6} \Rightarrow T = m g - \frac{m g}{3} = \frac{2 m g}{3}$ .

B: $T - m g sin θ = m \cdot \frac{g}{6} \Rightarrow sin θ = \frac{2}{3} - \frac{1}{6} = \frac{1}{2} \Rightarrow θ = 3 0^{\circ}$ .

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