Question

The WCB of a line is 236\(^{\circ}\) then its QB is

• A. N56\(^{\circ}\) E
• B. S 56\(^{\circ}\) E
• C. S 56\(^{\circ}\) W
• D. N 56\(^{\circ}\) W

Hint:

To quickly convert WCB to QB:
- NE (0-90): QB = WCB (e.g., 40\(^{\circ}\) $\rightarrow{}$ N 40\(^{\circ}\) E) - SE (90-180): QB = 180\(^{\circ}\) - WCB (e.g., 130\(^{\circ}\) $\rightarrow{}$ S 50\(^{\circ}\) E) - SW (180-270): QB = WCB - 180\(^{\circ}\) (e.g., 236\(^{\circ}\) $\rightarrow{}$ S 56\(^{\circ}\) W) - NW (270-360): QB = 360\(^{\circ}\) - WCB (e.g., 310\(^{\circ}\) $\rightarrow{}$ N 50\(^{\circ}\) W)

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need to convert a bearing from Whole Circle Bearing (WCB) format to Quadrantal Bearing (QB) format.

Step 2: Key Formula or Approach:
1. Determine the quadrant in which the WCB lies.
- 0\(^{\circ}\) to 90\(^{\circ}\): NE Quadrant - 90\(^{\circ}\) to 180\(^{\circ}\): SE Quadrant - 180\(^{\circ}\) to 270\(^{\circ}\): SW Quadrant - 270\(^{\circ}\) to 360\(^{\circ}\): NW Quadrant 2. Calculate the acute angle from the North or South line.

Step 3: Detailed Explanation:
The given WCB is 236\(^{\circ}\).
This value is between 180\(^{\circ}\) and 270\(^{\circ}\), so the line lies in the

South-West (SW) quadrant.
In the QB system, bearings in the SW quadrant are measured from the South line towards the West.
The angle from the South line (180\(^{\circ}\)) is calculated as:
\[ \text{Angle} = \text{WCB} - 180^{\circ} \] \[ \text{Angle} = 236^{\circ} - 180^{\circ} = 56^{\circ} \] Combining the quadrant and the angle, the QB is S 56\(^{\circ}\) W.

Step 4: Final Answer:
The Quadrantal Bearing is S 56\(^{\circ}\) W.
This corresponds to option (C).