$AB$ 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 $CD=7\text{m}$. From $D$, the angle of elevation of the point $A$ is $45^{\circ }$. Then the height of the pole is:

1. A
   $\frac{7\sqrt{3}}{2}(\sqrt{3}-1)\text{m}$
2. B
   $\frac{7\sqrt{3}}{2}(\sqrt{3}+1)\text{m}$
3. C
   $7\sqrt{3}(\sqrt{3}-1)\text{m}$
4. D
   $\frac{7\sqrt{3}}{\sqrt{3}-1}\text{m}$