Question

For a given traverse, latitudes and departures are calculated and it is found that sum of latitudes is equal to +2.1 m and the sum of departures is equal to -2.8 m. The length and bearing of the closing error, respectively, are

• A. 3.50 m and 53°7'48" NW
• B. 2.45 m and 53°7'48" NW
• C. 0.35 m and 53.13° SE
• D. 3.50 m and 53.13° SE

Hint:

To calculate the closing error, use the Pythagorean theorem, and to find the bearing, use the inverse tangent function. Adjust the sign based on the quadrant.

Answer 1

The Correct Option is A

Given: - Sum of latitudes = +2.1 m - Sum of departures = -2.8 m The closing error can be calculated using the Pythagorean theorem since the latitudes and departures form the two perpendicular components of the error: \[ \text{Closing error} = \sqrt{(\text{Sum of latitudes})^2 + (\text{Sum of departures})^2} \] Substituting the values: \[ \text{Closing error} = \sqrt{(2.1)^2 + (-2.8)^2} = \sqrt{4.41 + 7.84} = \sqrt{12.25} = 3.5 \, \text{m} \] The bearing of the closing error is calculated as the angle \( \theta \) with respect to the positive x-axis (eastward direction) using: \[ \theta = \tan^{-1}\left( \frac{\text{Sum of latitudes}}{\text{Sum of departures}} \right) = \tan^{-1}\left( \frac{2.1}{-2.8} \right) \] Since the sum of departures is negative and the sum of latitudes is positive, the angle lies in the NW quadrant. Calculating the angle: \[ \theta = \tan^{-1}\left( -\frac{2.1}{2.8} \right) = \tan^{-1}(-0.75) \approx -36.87^\circ \] Thus, the bearing is 53°7'48" NW. Therefore, the correct answer is option (A).
Final Answer: (A) 3.50 m and 53°7'48" NW