Question

If the fore bearing of a line AB is S$15^\circ40'$W, then the back bearing of line is

• A. $15^\circ40'$
• B. $364^\circ20'$
• C. S$15^\circ40'$E
• D. N$15^\circ40'$W

Hint:

To find the back bearing in the Quadrantal Bearing system, just flip the letters.
S becomes N, N becomes S. E becomes W, W becomes E. The angle stays the same.
Example: FB = S $30^\circ$ E $\implies$ BB = N $30^\circ$ W.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks to find the back bearing of a line, given its fore bearing in the Quadrantal Bearing (or Reduced Bearing) system. The provided answer key `15°40'` seems incomplete or formatted as a Whole Circle Bearing, let's derive the correct Quadrantal Bearing first.

Step 2: Key Formula or Approach:
The

Fore Bearing (FB) is the bearing of a line in the direction of the survey traverse. The

Back Bearing (BB) is the bearing of the same line in the opposite direction.
In the Quadrantal Bearing (QB) system, bearings are measured from either North or South towards East or West.
To find the back bearing from the fore bearing in the QB system, you simply:
1. Keep the numerical angle the same.
2. Reverse the cardinal directions (North becomes South, South becomes North, East becomes West, and West becomes East).

Step 3: Detailed Explanation:
The given Fore Bearing of line AB is S$15^\circ40'$W.
This means the line is in the South-West quadrant, making an angle of $15^\circ40'$ with the South direction, towards the West.
To find the Back Bearing of line BA, we apply the rule:
- Keep the angle: $15^\circ40'$.
- Reverse the South to North.
- Reverse the West to East.
So, the Back Bearing is N$15^\circ40'$E.
Now, let's evaluate the options. None of the options is N$15^\circ40'$E. There is a clear error in the question or the options. Let's analyze the given correct answer: "$15^\circ40'$". This looks like an angle in the Whole Circle Bearing (WCB) system, but the value is too small. Let's convert the Fore Bearing to WCB and see what we get.
FB in QB = S$15^\circ40'$W
This is in the third quadrant. WCB = $180^\circ + 15^\circ40' = 195^\circ40'$.
The Back Bearing in WCB is FB $\pm 180^\circ$.
BB in WCB = $195^\circ40' - 180^\circ = 15^\circ40'$.
This WCB of $15^\circ40'$ corresponds to a Quadrantal Bearing of N$15^\circ40'$E.
So, the calculation shows that the back bearing has a numerical value of $15^\circ40'$, and the provided answer key seems to have only stated this numerical part, omitting the cardinal directions. Option (A) matches this numerical value.

Step 4: Final Answer:
The back bearing is N$15^\circ40'$E. In the Whole Circle Bearing system, this is $15^\circ40'$. Given the options, the intended answer is the numerical value $15^\circ40'$.