3. As shown in Figure 2, quadrilateral $ABCD$ is a parallelogram, $\angle ADC=60^{\circ}, AD=2$, $AB=\sqrt{3}+1, M$ is the midpoint of side $AD$, and the angle bisector of $\angle BCD$ intersects segment $BM$ at point $K$. Find the value of $\angle CKB$.
3. Let $X$ be a point on line segment $AB$, satisfying $AX=1, XB=\sqrt{3}$.
From the problem, we know $\angle MAX=120^{\circ}$.
Using the cosine rule, we get $MX=\sqrt{3}$.
Since $MX=XB$, thus, $\angle MBX=\frac{1}{2} \angle MXA=15^{\circ}$.
Therefore, $\angle CBK=\angle ABC-\angle MBX=45^{\circ}$.
Hence, $\angle CKB=180^{\circ}-\angle BCK-\angle CBK$ $=180^{\circ}-60^{\circ}-45^{\circ}=75^{\circ}$.