Question

The QB of a line is S 30\(^{\circ}\) 20' E then its WCB is

• A. 149\(^{\circ}\) 40'
• B. 30\(^{\circ}\) 20'
• C. 329\(^{\circ}\) 40'
• D. 210\(^{\circ}\) 20'

Hint:

To quickly convert QB to WCB:
- N...E: WCB = QB (e.g., N 40\(^{\circ}\) E $\rightarrow{}$ 40\(^{\circ}\)) - S...E: WCB = 180\(^{\circ}\) - QB (e.g., S 30\(^{\circ}\) E $\rightarrow{}$ 150\(^{\circ}\)) - S...W: WCB = 180\(^{\circ}\) + QB (e.g., S 50\(^{\circ}\) W $\rightarrow{}$ 230\(^{\circ}\)) - N...W: WCB = 360\(^{\circ}\) - QB (e.g., N 50\(^{\circ}\) W $\rightarrow{}$ 310\(^{\circ}\))

Answer 1

The Correct Option is A

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

Step 2: Key Formula or Approach:
1. Identify the quadrant from the QB notation.
2. Apply the appropriate conversion rule based on the quadrant.
WCB is always measured clockwise from the North line (0\(^{\circ}\)).

Step 3: Detailed Explanation:
The given QB is S 30\(^{\circ}\) 20' E.
This indicates the line is in the

South-East (SE) quadrant.
The bearing is measured from the South line (180\(^{\circ}\)) towards the East.
To find the WCB, we must subtract this angle from 180\(^{\circ}\).
\[ \text{WCB} = 180^{\circ} - 30^{\circ} 20' \] To perform the subtraction, we can rewrite 180\(^{\circ}\) as 179\(^{\circ}\) 60'.
\[ \text{WCB} = 179^{\circ} 60' - 30^{\circ} 20' \] \[ \text{WCB} = (179 - 30)^{\circ} (60 - 20)' \] \[ \text{WCB} = 149^{\circ} 40' \]

Step 4: Final Answer:
The Whole Circle Bearing is 149\(^{\circ}\) 40'.
This corresponds to option (A).