What is the angle needed to solve this right triangle?

[nmnna]

Homework Statement

In figure $\angle{ABC} = 90^\circ = \angle{BCD}, \ \angle{ACB} = 41.45^\circ, \ \angle{CBD} = 32.73^\circ, \ BC = 10$cm. Calculate $AB, \ CD$ and $\angle{AEB}$

Relevant Equations

$\tan{\alpha} = \frac{opposite \ side}{adjacent \ side}$

The Figure

My Attempt at Solution

$\tan{ACB} = \frac{AB}{BC}, \ \tan41.45^\circ = \frac{AB}{10} \Rightarrow AB = 10\tan45.41^\circ \approx 8.83$cm
Similarly
$\tan{CBD} = \frac{CD}{BC}, \ \tan32.73^\circ = \frac{CD}{10} \Rightarrow CD = 10\tan32.73^\circ \approx 6.43$cm
After this I checked the answer in my textbook, and instead of 6.43cm the answer for $CD$ was 4.40cm.
I thought that there was a typo in the problem, so instead of $32.73^\circ$, I tried $23.73^\circ$, and surprisingly the answer matches with the one in the textbook.
So I'd like to know if it really is a typo or my solution is wrong.
And I can't find the angle required in the problem, so I'd be grateful if you give me some hints for finding this angle.

 

[PeroK]

Homework Statement:: In figure ##\angle{ABC} = 90^\circ = \angle{BCD}, \ \angle{BCD} = 41.45^\circ,

You have angle $BCD$ twice there. Please clarify.

 

[nmnna]

You have angle $BCD$ twice there. Please clarify.

It should be $ACB$

 

[PeroK]

So I'd like to know if it really is a typo or my solution is wrong.
And I can't find the angle required in the problem, so I'd be grateful if you give me some hints for finding this angle.

Looks like a typo.

A hint to find the angle $AEB$. First extend the line $AB$ to a new point, $F$, so that $AFD$ is a right angle. Then you have another angle equal to $AEB$.

 

[FactChecker]

Your work is correct starting with the numbers you were given.