$AB$ is a vertical pole with $B$ at the ground level and $A$ at the top. $A$ man finds that the angle of elevation of the point $A$ from a certain point $C$ on the ground is $60^{\circ}$. He moves away from the pole along the line $BC$ to a point $D$ such that $CD = 7 \ m$. From $D$,the angle of elevation of the point $A$ is $45^{\circ}$. Then the height of the pole is

• A
  $\frac{7\sqrt{3}}{2(\sqrt{3}-1)}$
• B
  $\frac{7\sqrt{3}}{2}(\sqrt{3}+1)$
• C
  $\frac{7\sqrt{3}}{2}(\sqrt{3}-1)$
• D
  $\frac{7\sqrt{3}}{2(\sqrt{3}+1)}$