Answer
We can rank the values of $\phi$ according to the magnitude of the angular acceleration of the bar:
$90^{\circ} \gt 70^{\circ} = 110^{\circ}$
Work Step by Step
The torque due to $F_1$ tends to make the bar rotate counterclockwise.
The torque due to $F_2$ also tends to make the bar rotate counterclockwise.
To maximize the angular acceleration, we should maximize the torque.
The expression for the torque due to $F_2$ is: $~~\tau_2 = r_2~F_2~sin~\phi$
The torque is a maximum when $\phi = 90^{\circ}$
Since $~~sin~70^{\circ} = sin~110^{\circ},~~$ the torque is equal at both of these angles of $\phi$
We can rank the values of $\phi$ according to the magnitude of the angular acceleration of the bar:
$90^{\circ} \gt 70^{\circ} = 110^{\circ}$
