Answer
$\theta=142^{\circ}$
Work Step by Step
$\begin{aligned} & \vec{\gamma}_{B C}=\{6 \mathrm{i}+4 \mathrm{j}-2 \hat{\mathrm{k}}\} \mathrm{ft} \\ & \vec{\gamma}_{B A}=\{-3 \hat{\mathrm{i}}\} \mathrm{ft} \\ & \theta=\cos ^{-1}\left(\frac{\vec{\gamma}_{B C} \cdot \vec{\gamma}_{B A}}{\left|\vec{\gamma}_{B C}\right|\left|\vec{\gamma}_{B A}\right|}\right)=\cos ^{-1}\left(\frac{-18}{22.45}\right) \\ & \theta=143^{\circ}\end{aligned}$
