Question

The Whole Circle Bearing (WCB) of line AB is \(150^{\circ}\). The length of AB is 100 m. The latitude and departure values, respectively, for this line are:

• A. +86.60 ; +50.00
• B. +86.60 ; -50.00
• C. -86.60 ; +50.00
• D. -86.60 ; -50.00

Hint:

Convert the WCB to a reduced quadrantal bearing to fix the signs, then use L cos and L sin for the magnitudes of latitude and departure.

Answer 1

The Correct Option is C

An easier route to the same answer is to first convert the whole circle bearing into a reduced (quadrantal) bearing, since quadrantal bearings make the sign of each component obvious by inspection instead of relying purely on the sign of sine and cosine.

The WCB of AB is $150^{\circ}$. Since this falls between $90^{\circ}$ and $180^{\circ}$, it belongs to the second quadrant, measured from south. Convert it to a reduced bearing using:

\[ \text{Reduced bearing} = 180^{\circ} - \text{WCB} = 180^{\circ} - 150^{\circ} = 30^{\circ} \]

So the reduced bearing is S $30^{\circ}$ E, meaning the line runs 30 degrees east of due south. This directly tells us the direction sense: south gives a negative latitude, and east gives a positive departure.

Now find the magnitudes using the reduced bearing angle of $30^{\circ}$ with basic trigonometry, since the length of AB is 100 m:

\[ |\text{Latitude}| = L \cos(30^{\circ}) = 100 \times 0.8660 = 86.60 \text{ m} \] \[ |\text{Departure}| = L \sin(30^{\circ}) = 100 \times 0.500 = 50.00 \text{ m} \]

Applying the direction signs identified from the reduced bearing S $30^{\circ}$ E (south is negative latitude, east is positive departure):

\[ \text{Latitude} = -86.60 \text{ m}, \qquad \text{Departure} = +50.00 \text{ m} \]

This matches option (C). The reduced-bearing method is a useful check on the direct $L\cos\theta$, $L\sin\theta$ calculation, since it separates the magnitude (always from a $0^{\circ}$-$90^{\circ}$ acute angle) from the sign (fixed purely by which quadrant the WCB falls in).

\[ \boxed{\text{Latitude} = -86.60 \text{ m}, \ \text{Departure} = +50.00 \text{ m}} \]