Question

i think the answer is the first one. $\frac{33 \sqrt{14} }{2}$am I right?

Answer 1

Given:

$AB=11\sqrt[]{7}$

Consider right triangle ABE.

Applying trigonometric property in triangle ABE,

$\begin{gathered} \tan 60^{\circ}=\frac{opposite\text{ side }}{adjacent\text{ side}} \\ \tan 60^{\circ}=\frac{AB}{BE} \\ \sqrt[]{3}=\frac{11\sqrt[]{7}}{BE} \\ BE=\frac{11\sqrt[]{7}}{\sqrt[]{3}} \end{gathered}$

Consider right triangle BED.

Applying trigonometric property in triangle BED,

$\begin{gathered} \cos 45^{\circ}=\frac{adjacent\text{ side}}{hypotenuse} \\ \frac{1}{\sqrt[]{2}}=\frac{BE}{BD} \\ \frac{1}{\sqrt[]{2}}=\frac{\frac{11\sqrt[]{7}}{\sqrt[]{3}}}{BD} \\ BD=\frac{11\sqrt[]{7}\times\sqrt[]{2}}{\sqrt[]{3}} \\ BD=\frac{11\sqrt[]{14}}{\sqrt[]{3}} \end{gathered}$

Consider right triangle BDC. Applying trigonometric property in triangle BDC,

$\begin{gathered} \tan 30^{\circ}=\frac{opposite\text{ side}}{\text{adjacent side}} \\ \frac{1}{\sqrt[]{3}}=\frac{BD}{CD} \\ \frac{1}{\sqrt[]{3}}=\frac{\frac{11\sqrt[]{14}}{\sqrt[]{3}}}{CD} \\ CD=11\sqrt[]{14} \end{gathered}$

Therefore,

$CD=11\sqrt[]{14}$