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Chapter 1 of 12 • Page 1 of 248🔒 Protected PDF • Watermarked
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CivilAdvanced Survey
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Calculate length and bearing of line. When it has a latitude and departure are -50m.

A

70.72 and S45°E

B

70.72 and S45°W

C

70.72 and 225°

D

Both B and C

Correct Answer

⚙️ TE • Technical Direct FormulaCivilAdvanced Survey
Option D

Both B and C

Quick Summary:

Given: Latitude (L) = -50 m, Departure (D) = -50 m

📐MAMath SolutionDirect Formula
📋 Given

Latitude (L) = -50 m, Departure (D) = -50 m

🔢 Formula Used

l=L2+D2,θ=tan⁡−1∣DL∣l = \sqrt{L^2 + D^2}, \quad \theta = \tan^{-1}\left|\frac{D}{L}\right|l=L2+D2​,θ=tan−1​LD​​

📊 Diagram / Illustration
N (+L) S (-L) W (-D) E (+D)
L=−50L = -50L=−50 m
D=−50D = -50D=−50 m
l=70.72l = 70.72l=70.72 m
45∘45^{\circ}45∘
WCB=225∘\text{WCB} = 225^{\circ}WCB=225∘
🔢 Step-by-Step Solution
1

Calculate Length of Line

The length (lll) of a line is calculated using the Pythagorean theorem with given Latitude (L=−50 mL = -50\text{ m}L=−50 m) and Departure (D=−50 mD = -50\text{ m}D=−50 m).

l=L2+D2=(−50)2+(−50)2=5000≈70.7107 m≈70.72 ml = \sqrt{L^2 + D^2} = \sqrt{(-50)^2 + (-50)^2} = \sqrt{5000} \approx 70.7107 \text{ m} \approx 70.72 \text{ m}l=L2+D2​=(−50)2+(−50)2​=5000​≈70.7107 m≈70.72 m

2

Calculate Reduced Bearing (RB / Quadrantal Bearing)

Find the acute angle θ\thetaθ with the meridian: tan⁡θ=∣DL∣\tan \theta = \left|\frac{D}{L}\right|tanθ=​LD​​.

tan⁡θ=∣−50−50∣=1  ⟹  θ=45°\tan \theta = \left|\frac{-50}{-50}\right| = 1 \implies \theta = 45°tanθ=​−50−50​​=1⟹θ=45°

3

Determine Quadrant and Reduced Bearing

Since both Latitude (LLL) and Departure (DDD) are negative, the line lies in the 3rd quadrant (South-West quadrant). Therefore, the Reduced Bearing is S45°W\text{S}45°\text{W}S45°W.

RB=S45°W\text{RB} = \text{S}45°\text{W}RB=S45°W

4

Calculate Whole Circle Bearing (WCB)

In the 3rd quadrant, Whole Circle Bearing is calculated as WCB=180°+θ\text{WCB} = 180° + \thetaWCB=180°+θ.

WCB=180°+45°=225°\text{WCB} = 180° + 45° = 225°WCB=180°+45°=225°

5

Compare Options

Option B gives the length as 70.7270.7270.72 and bearing in RB as S45°W\text{S}45°\text{W}S45°W. Option C gives length as 70.7270.7270.72 and bearing in WCB as 225°225°225°. Since both expressions for bearing are valid, Option D ('Both B and C') is the correct choice.

Bearing=S45°W=225°\text{Bearing} = \text{S}45°\text{W} = 225°Bearing=S45°W=225°

✅

D is correct because both S45°W\text{S}45°\text{W}S45°W (Reduced Bearing) and 225°225°225° (Whole Circle Bearing) correctly represent the bearing of the line with length 70.72 m70.72\text{ m}70.72 m.

Core Concepts Used
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Latitude and Departure Reduced Bearing vs. Whole Circle Bearing Traverse Computation
💡 EXAM TIP

Always check whether options represent bearing in Whole Circle Bearing (WCB) or Quadrantal/Reduced Bearing (RB), as both system notations are standard in surveying questions.

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