Question

I think the answer is the first one.
`(33 √(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)=(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}}=(BE)/(BD) \\ \frac{1}{\sqrt[]{2}}=\frac{\frac{11\sqrt[]{7}}{\sqrt[]{3}}}{BD} \\ BD=\frac{11\sqrt[]{7}*\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}}=(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}`