Answer
$v_B=12.6058in/s$
$\theta=65.657^{\circ}$
Work Step by Step
The required velocity can be determined as follows:
We know that
$\vec{r_{B/G}}=\frac{1.5}{sin45}\hat i=2.1213\hat i$
and the velocity at point G is $\vec{v_{G}}=-6cos30\hat i+6sin30\hat j=-5.196\hat i+3\hat j$
Now, $\vec{v_B}=\vec{v_G}+\vec{\omega}\times \vec{r_{B/G}}$
We plug in the known values to obtain:
$\vec{v_B}=(-5.196\hat i+3\hat j)+(4\hat k)\times(2.1213\hat i)$
$\implies \vec{v_B}=-5.196\hat i+11.4852\hat j$
The magnitude of the velocity is given as
$v_B=\sqrt{(v_{Bx})^2+(v_{By})^2}$
We plug in the known values to obtain:
$v_B=\sqrt{(-5.196)^2+(11.4852)^2}$
$\implies v_B=12.6058in/s$
and the angle can be determined as
$\theta=tan^{-1}|\frac{v_{By}}{v_{Bx}}|$
$\implies \theta=tan^{-1}|\frac{11.4852}{-5.196}|=65.657^{\circ}$
