In summary: No, the angle of rotation is 0 degrees. What does that mean for the orientation of the element?It means that the element must be rotated so that the normal stress is at (\sigma_{y},+\tau_{xy}) instead of (\sigma_{y},-\tau_{xy}). Does that make sense?Yes, that makes sense.How would I go about finding the orientation of the element if I wanted it to be at (\sigma_{y},+\tau_{xy})?If you wanted the element to be at (\sigma_{y},+\tau_{xy}), you'd find the x-coordinate and y-coordinate of (\
Your Mohr's circle is almost complete. Try not to forget to draw the sigma-tau reference before you draw a circle. Then you will see that the circle is on the left side of the reference (Uniaxial compression). Note that the right part of the circle touches the origin. I think now the problem solved. If you want stresses at X’Y’, just rotates the horizontal line through the angle you want (it doesn’t matter whether the angle is given since, generally, we often need the principle stresses and maximum shear stress which are on the horizontal and vertical lines)