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

Step 1: Recall the definitions of latitude and departure.
In traverse surveying, for a line of length \(L\) with Whole Circle Bearing \(\theta\) (measured clockwise from north), the latitude (the north-south component) and departure (the east-west component) are given by
\[ \text{Latitude} = L \cos\theta, \qquad \text{Departure} = L \sin\theta \]
Latitude is taken positive when northing and negative when southing; departure is taken positive when easting and negative when westing.

Step 2: Identify the quadrant of the bearing.
The WCB is measured clockwise from north: \(0^{\circ}\) to \(90^{\circ}\) is the NE quadrant, \(90^{\circ}\) to \(180^{\circ}\) is the SE quadrant, \(180^{\circ}\) to \(270^{\circ}\) is the SW quadrant, and \(270^{\circ}\) to \(360^{\circ}\) is the NW quadrant.
Here \(\theta = 150^{\circ}\), which lies between \(90^{\circ}\) and \(180^{\circ}\), so line AB runs in the SE quadrant. That means it heads south (negative latitude) and east (positive departure).

Step 3: Compute latitude and departure.
With \(L = 100\) m and \(\theta = 150^{\circ}\):
\[ \cos(150^{\circ}) = -\cos(30^{\circ}) = -0.8660 \]
\[ \sin(150^{\circ}) = \sin(30^{\circ}) = 0.5000 \]
So:
\[ \text{Latitude} = 100 \times (-0.8660) = -86.60 \text{ m} \]
\[ \text{Departure} = 100 \times (0.5000) = +50.00 \text{ m} \]
The negative latitude confirms the southing direction, and the positive departure confirms the easting direction, matching the SE quadrant identified in Step 2.

Step 4: Final Answer:
The latitude of AB is -86.60 m and the departure is +50.00 m. \[ \boxed{\text{Latitude} = -86.60 \text{ m}, \ \text{Departure} = +50.00 \text{ m}} \]