Complete this table in order to calculate, with reasons the size of B2.

Here is the completed table with the calculations and reasons:

Statement	Reason
$\hat{C} = \mathbf{40^\circ}$	Given
$\hat{B}_2 = \mathbf{\hat{F}_3}$	$\angle s$ opp. equal sides (in $\triangle BFC$)
$2\hat{F}_3 + 40^\circ = 180^\circ$	Sum of angles in $\triangle BFC$
$2\hat{F}_3 = \mathbf{140^\circ}$	Calculation
$\therefore \hat{F}_3 = \mathbf{70^\circ}$	Calculation
But $\hat{F}_3 = \mathbf{\hat{B}_1}$	alt. $\angle s$ and AB $\parallel$ DC (transversal BF)
and $\hat{B}_1 = \mathbf{70^\circ}$	Substitution
$\hat{F}_2 + \hat{B}_1 + \hat{E}_2 = 180^\circ$	Sum of angles in $\triangle EFB$
$\mathbf{\hat{E}_2 = \hat{B}_1}$	$\angle s$ opp. equal sides (in $\triangle EFB$, since EF=BF)
$\therefore \hat{F}_2 = \mathbf{40^\circ}$	Calculation

The final value is $\hat{F}_2 = 40^\circ$.