The locations where the passneger travels can be shown as,
The position of passenger at city A can be expressed as,
[tex]\begin{gathered} \vec{A}=(175km)\cos 30^{\circ}\hat{i}+(175km)\sin 30^{\circ}\hat{j} \\ =(175\text{ km)(}0.866)\hat{i}+(175\text{ km)(}0.5)\hat{j} \\ =151.55\text{ km }\hat{\text{i}}+87.5\text{ km }\hat{\text{j}} \end{gathered}[/tex]The position of passenger at city B can be expressed as,
[tex]\begin{gathered} \vec{B}=((151.55km)-(153km)\sin 20^{\circ})\hat{i}+(87.5km+(153km)\cos 20^{\circ})\hat{j} \\ =((151.55km)-(153\text{ km)(}0.342))\hat{i}+(87.5km+(153\text{ km)(}0.94))\hat{j} \\ =(151.55km-52.33\text{ km) }\hat{\text{i}}+(87.5km+143.82\text{ km) }\hat{\text{j}} \\ =99.22\text{ km }\hat{\text{i}}+231.32\text{ km}\hat{\text{j}} \end{gathered}[/tex]The position of passenger at city C can be expressed as,
[tex]\begin{gathered} \vec{C}=(99.22km-195km)\hat{i}+231.32\operatorname{km}\hat{j}_{} \\ =-95.78\text{ }\hat{\text{i}}+231.32\text{ km}\hat{\text{j}} \end{gathered}[/tex]Therefore, the magnitude of distance of city C can be calculated as,
[tex]\begin{gathered} C=\sqrt[]{(-95.78km)^2+(231.32km)^2} \\ =\sqrt[]{9173.81km^2+53508.94km^2} \\ =252.3\text{ km} \end{gathered}[/tex]Thus, the distance of city C from point O is 252.3 km.
The direction of location is calculated as,
[tex]\begin{gathered} \cos \theta=\frac{-95.78^{}}{231.32} \\ \theta=-65.5^{\circ} \end{gathered}[/tex]Thus, the direction of city C from point O is -65.5 degree.
