Answer
$\begin{aligned} & \theta=74.4^{\circ} \\ & \phi=55.4^{\circ}\end{aligned}$
Work Step by Step
$\begin{array}{ll}\mathbf{r}_{A C}=[-2 \mathbf{i}-4 \mathbf{j}+1 \mathbf{k}] \mathrm{m} ; & r_{A C}=4.58 \mathrm{~m} \\ \mathbf{r}_{A B}=[1.5 \mathbf{i}-4 \mathbf{j}+3 \mathbf{k}] \mathrm{m} ; & r_{A B}=5.22 \mathrm{~m} \\ \mathbf{r}_{A O}=\{-4 \mathbf{j}-3 \mathbf{k}\} \mathrm{m} ; & r_{A O}=5.00 \mathrm{~m} \\ \mathbf{r}_{A B} \cdot \mathbf{r}_{A O}=(1.5)(0)+(-4)(-4)+(3)(-3)=7 \\ \theta=\cos ^{-1}\left(\frac{\mathbf{r}_{A B^*} \mathbf{r}_{A O}}{r_{A B} r_{A O}}\right) \\ =\cos ^{-1}\left(\frac{7}{5.22(5.00)}\right)=74.4^{\circ} \\ \mathbf{r}_{A C} \cdot \mathbf{r}_{A O}=(-2)(0)+(-4)(-4)+(1)(-3)=13 \\ \phi=\cos ^{-1}\left(\frac{\mathbf{r}_{A C} \cdot \mathbf{r}_{A O}}{r_{A C} r_{A O}}\right) \\ =\cos ^{-1}\left(\frac{13}{4.58(5.00)}\right)=55.4^{\circ}\end{array}$
