Answer
$\omega_{AB}=3.46 rad/s$
$\omega_{BC}=2.31rad/s$
Work Step by Step
The required angular velocity can be determined as follows:
We know that
$\vec{v_B}=\vec{v_C}+\vec{\omega_{BC}}\times \vec{r_{B/C}}~$[eq(1)]
The velocity of $C$ is given as $\vec{v_C}=6\hat j$
and $\vec{r_{B/C}}=-3cos30\hat j+3sin30\hat j=-2.598\hat i+1.5\hat j$
The velocity of $B$ is
$\vec{\omega_B}=\vec{\omega_{AB}}\times \vec r_{AB}$
$\implies \vec{v_B}=(\omega_{AB})\hat k\times (1\hat j)=-\omega_{AB}\hat i$
We plug in the known values in eq(1) to obtain:
$-\omega_{AB}\hat i=6\hat j+(\omega_{BC})\hat k(-2.598\hat i+1.5\hat j)$
$\implies \omega_{AB}\hat i=-1.5\omega_{BC}\hat i+(6-2.598\omega_{BC})\hat j~$[eq(2)]
Comparing $i$ components on both sides of eq(2)
$-\omega_{AB}=-1.5\omega_{BC}$
$\implies \omega_{AB}=1.5\omega_{BC}~$[eq(3)]
Comparing $j$ components of eq(2) on both sides
$0=6-2.598\omega_{BC}$
This simplifies to:
$\omega_{BC}=2.3095rad/s$
We plug in this value in eq(3) to obtain:
$\omega_{AB}=1.5(2.3095)=3.464 rad/s$
