Find pressure at point A in the given figure.........

Let the adjacent point of sides l2 and l3 be B and that of sides l1 and l2 be D and the free surface of liquid at open end of pipe be E.Pressure at E = atmospheric pressure = Patm.Since fluid is static, pressure difference is due to height of fluid.Pressure at C = Pressure at E + \rhogh1,\rho = density of fluid in pipe and h1 = height of fluid between E and C = l1 sin\theta1.i.e. PC = Patm + \rhogl1 sin\theta1.Pressure at C = Pressure at B + \rhogh2 , h2 = l2 sin\theta2.i.e. PC = PB + \rhogl2 sin\theta2 \Rightarrow PB = PC – \rhogl2 sin\theta2 = Patm + \rhogl1 sin\theta1 – \rhogl2 sin\theta2Similarly, PA = PB + \rhogl3 sin\theta3 = Patm + \rhogl1 sin\theta1 – \rhogl2 sin\theta2 + \rhogl3 sin\theta3PA = \rhog(l1 sin\theta1 – l2 sin\theta2 + l3 sin\theta3) = pressure at A .