My friend sent me that question and said "another Carnot theorem" is used to solve this question but i couldnt find that theorem. Can you help me?
Additional explanation: $$ \widehat{ABD} = 30^{\circ} $$ $$ \widehat{DBC} = 1^{\circ} $$ $$ \widehat{ACD} = 89^{\circ} $$ $$ \widehat{BAD} = \widehat{DAC} $$ $$ x(\widehat{BCD}) = ? $$

You should use Ceva's theorem (in trigonometric form -- http://www.cut-the-knot.org/triangle/TrigCeva.shtml)
We have
$\frac{\sin 1^\circ\sin 89^\circ}{\sin 30^\circ\sin x}=1$, i. e.
$sin x = \frac{\sin 1^\circ \cos 1^\circ}{\sin 30^\circ} = 2 \sin 1^\circ \cos 1^\circ=sin 2^\circ$
$x=2^\circ$