Question

A delivery agent travels from $R$ to $P$ along straight-line paths $RC$, $CA$, $AB$, and $BP$, each of length $5\,$km. The whole circle bearings (clockwise from North) are: $RC=120^\circ$, $CA=0^\circ$, $AB=90^\circ$, $BP=240^\circ$. If the latitude $(L)$ and departure $(D)$ of $R$ are $(0,0)$ km, find the latitude and departure of $P$ (rounded to one decimal place). \includegraphics[width=0.75\linewidth]{image666.png}

• A. $L=2.5;\; D=5.0$
• B. $L=0.0;\; D=5.0$
• C. $L=5.0;\; D=2.5$
• D. $L=0.0;\; D=0.0$

Hint:

For whole circle bearings (measured clockwise from North), use $L=r\cos\theta$ and $D=r\sin\theta$. Positive $L$ points North and positive $D$ points East. Angles in opposite quadrants often cancel nicely.

Answer 1

The Correct Option is B

Step 1 (Components for each traverse leg): In whole circle bearings, latitude is the North–South component and departure is the East–West component. For a leg of length $r=5$ km at bearing $\theta$, \[ L=r\cos\theta, D=r\sin\theta, \] taking $+\!L$ as North, $-\!L$ as South, $+\!D$ as East, and $-\!D$ as West.
Step 2 (Compute each leg): \[ \begin{aligned} RC:~\theta&=120^\circ &\Rightarrow&~ L_{RC}=5\cos120^\circ=5(-\tfrac12)=-2.5, D_{RC}=5\sin120^\circ=5(\tfrac{\sqrt3}{2})\approx +4.33.
CA:~\theta&=0^\circ &\Rightarrow&~ L_{CA}=5\cos0^\circ=+5.0,\ \ \ \, D_{CA}=5\sin0^\circ=0.
AB:~\theta&=90^\circ &\Rightarrow&~ L_{AB}=5\cos90^\circ=0,\ \ \ \, D_{AB}=5\sin90^\circ=+5.0.
BP:~\theta&=240^\circ &\Rightarrow&~ L_{BP}=5\cos240^\circ=5(-\tfrac12)=-2.5, D_{BP}=5\sin240^\circ=5(-\tfrac{\sqrt3}{2})\approx -4.33. \end{aligned} \] Step 3 (Net latitude and departure from $R$ to $P$): \[ \begin{aligned} L_P&=\sum L_i=(-2.5)+5.0+0+(-2.5)=0.0~\text{km},
D_P&=\sum D_i=(+4.33)+0+5.0+(-4.33)=+5.0~\text{km}. \end{aligned} \] \[ \boxed{L=0.0~\text{km}, D=5.0~\text{km}} \]