1. A particle of mass 0.5 kg is at rest on a rough plane which is inclined at $35^{\circ}$ to the horizontal.  The particle is being acted upon by a force of 6 N, directed at an angle of $\theta ^{\circ}$ to the plane.  The line of action of the force is in the same vertical plane as the line of greatest slope of the inclined plane.

Given that the coefficient of friction between the particle and the plane is 0.4, and that the particle is on the point of slipping up the plane, find the value of $\theta$.

You may use the following trigonometric identity in your working without proof:

$6cos \theta -2.4sin \theta \equiv\frac{6\sqrt{29}}{5}cos( \theta +{tan}^{-1}(0.4))$