Question

If the departure and latitude of a line are \(70\ \text{m}\) and \(-130\ \text{m}\), respectively, then the whole circle bearing (WCB) of the line in degrees is

• A. 28
• B. 62
• C. 152
• D. 208

Hint:

Use signs to pick the quadrant: \(+\)dep/\(+\)lat\(\to\)NE, \(+\)dep/\(-\)lat\(\to\)SE, \(-\)dep/\(-\)lat\(\to\)SW, \(-\)dep/\(+\)lat\(\to\)NW. Then: NE \(\Rightarrow\) WCB \(=\theta\); SE \(\Rightarrow\) \(180^\circ-\theta\); SW \(\Rightarrow\) \(180^\circ+\theta\); NW \(\Rightarrow\) \(360^\circ-\theta\).

Answer 1

The Correct Option is C

Step 1: Identify the quadrant using signs.
Departure \(+70\) m \(\Rightarrow\) \(\text{East}\). Latitude \(-130\) m \(\Rightarrow\) \(\text{South}\). Hence the line lies in the \(\text{SE}\) quadrant (reduced bearing \(= S\,\theta\,E\)).

Step 2: Compute the reduced bearing angle.
\[ \tan\theta = \frac{|\text{dep}|}{|\text{lat}|}=\frac{70}{130}=0.53846 \Rightarrow \theta \approx \arctan(0.53846)\approx 28.3^\circ. \]

Step 3: Convert reduced bearing to whole circle bearing.
For the SE quadrant, \(\text{WCB} = 180^\circ - \theta\). Thus, \[ \text{WCB} \approx 180^\circ - 28.3^\circ = 151.7^\circ \approx \boxed{152^\circ}. \]