Does pushing a door at an angle require more force than pushing it straight back? We have door in our office a step higher than where we stand hence the handle too is slightly higher than the norm. And I constantly feel that we exert more force to open that door than the other doors in the office, as we always tend to push it at an upward angle. My assumption is that this angle is not productive and hence we exert more force unknowingly.
Is my assumptiom right? In which case would like to understand the physics behind it.
 A: Because the door is hinged, its motion is in the horizontal direction. The force applied to open it, if not applied in the horizontal direction , is a vector, with a horizontal component and a vertical component, thus at an angle , the total force must be larger, as only the horizontal component can be used in swinging the door.(vector analysis)
A: When you open a door, you apply a torque to the door. The more the torque, the faster it opens up. The measure of this torque is given by $$ \large\tau= R\times F $$ Where $\tau$ is the torque and $F$ is the force you apply and $R$ is where you apply it relative to the hinge of the door. The $\times$ symbolizes cross product which tells us it also depends on the angle between R anf F.
We can write it as $|\large\tau |= |R|.|F|\sin \theta$. The more angle between where you apply the force and the force, the more the torque. Since maximum value of sine is 1 at $\theta = 90^0$, we conclude maximum torque is when we are applying force to the door at $90^0$ to the surface.  
Also note that if you apply the same force nearer to the edge (R goes down) torque will go down and it gets harder to push the door. And it is theoretically impossible to push the door if we apply force directly at the hinge.(R=0)
