Question

If the length and reduced bearing of a line are 100 m and S60°W respectively, then the departure is

• A. 50 m
• B. 70.7 m
• C. 86.6 m
• D. 100 m

Hint:

Remember the formulas for Latitude and Departure:
- Latitude = Length $\times$ cos($\theta$) (L-L-C)
- Departure = Dength $\times$ sin($\theta$) (Wait, this doesn't work. Use D = L sin($\theta$))
Latitude is the North-South component. Departure is the East-West component.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks to calculate the departure of a survey line, given its length and its bearing in the Quadrantal Bearing (or Reduced Bearing) system.

Step 2: Key Formula or Approach:
In surveying, the coordinates of a line are resolved into two components:
-

Latitude (L): The projection of the line onto the North-South meridian. It is calculated as $L = \text{length} \times \cos(\theta)$. Latitude is positive for Northings and negative for Southings.
-

Departure (D): The projection of the line onto the East-West line. It is calculated as $D = \text{length} \times \sin(\theta)$. Departure is positive for Eastings and negative for Westings.
Here, $\theta$ is the reduced bearing angle.

Step 3: Detailed Explanation:
We are given:
- Length of the line = 100 m
- Reduced Bearing = S$60^\circ$W
This means the angle $\theta$ with the South (North-South meridian) is $60^\circ$.
Calculate the departure:
\[ D = \text{length} \times \sin(\theta) \] \[ D = 100 \text{ m} \times \sin(60^\circ) \] We know that $\sin(60^\circ) = \frac{\sqrt{3}}{2} \approx 0.866$.
\[ D = 100 \times 0.866 = 86.6 \text{ m} \] Since the bearing is in the South-

West quadrant, the departure is negative (a Westing). However, the question asks for the magnitude of the departure.

Step 4: Final Answer:
The departure of the line is 86.6 m.