Pressure difference between containers A and B

In an inclined limb only the vertical height of a column produces pressure: $h = L\sin\theta$.

Assumed reading of the figure: liquid of $S = 0.8$ stands 550 mm vertically above the mercury surface in A's limb, and water stands 300 mm vertically above the mercury surface in B's limb. The mercury surface lies 130 mm along the 45° limb on the A side and 230 mm along the 60° limb on the B side, measured from the common bottom of the U-tube.

Vertical heights of the mercury surfaces

$$\begin{aligned} h_A &= 0.130\sin 45^\circ = 0.0919\ \text{m} \\ h_B &= 0.230\sin 60^\circ = 0.1992\ \text{m} \\ \Delta h &= 0.1992 - 0.0919 = 0.1073\ \text{m} \end{aligned}$$

So mercury on the B side stands 0.1073 m higher.

Pressure balance at the level of the lower (A-side) mercury surface

$$p_A + \rho_{0.8}g(0.55) = p_B + \rho_w g(0.30) + \rho_{Hg}g(0.1073)$$ $$\begin{aligned} p_A - p_B &= 9.81\,[0.30 + 13.6(0.1073) - 0.8(0.55)] \\ &= 2.943 + 14.310 - 4.316 \\ &= 12.94\ \text{kN/m}^2 \end{aligned}$$

Answer: $p_A - p_B \approx 12.94$ kN/m² (container A at the higher pressure), for the reading above. Use the same $L\sin\theta$ method if the figure places the lengths differently.