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CivilAdvanced Survey
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How many latitude and departure for line when it has bearing S50┬░E and length 50m?

A

A) L=32.13┬а(N)L = 32.13\text{ (N)}L=32.13┬а(N), D=38.30┬а(W)D = 38.30\text{ (W)}D=38.30┬а(W)

B

B) L=32.13┬а(N)L = 32.13\text{ (N)}L=32.13┬а(N), D=38.30┬а(E)D = 38.30\text{ (E)}D=38.30┬а(E)

C

C) L=32.13┬а(S)L = 32.13\text{ (S)}L=32.13┬а(S), D=38.30┬а(E)D = 38.30\text{ (E)}D=38.30┬а(E)

D

D) L=32.13┬а(S)L = 32.13\text{ (S)}L=32.13┬а(S), D=38.30┬а(W)D = 38.30\text{ (W)}D=38.30┬а(W)

Correct Answer

тЪЩя╕П TE тАв Technical Concept & PrincipleCivilAdvanced Survey
Option C

L=32.13┬а(S)L = 32.13\text{ (S)}L=32.13┬а(S), D=38.30┬а(E)D = 38.30\text{ (E)}D=38.30┬а(E)

Quick Summary:

The latitude (LLL) and departure (DDD) of a line are its orthogonal components along the North-South and East-West meridian directions, respectively. For a line of length lll and reduced bearing ╬╕\theta╬╕ in the SE quadrant (S ╬╕\theta╬╕ E), the latitude is directed South (lcosтБб╬╕l \cos \thetalcos╬╕) and the departure is directed East (lsinтБб╬╕l \sin \thetalsin╬╕).

тЪЩя╕ПTETechnical SolutionConcept & Principle
ЁЯТб Explanation

The latitude (LLL) and departure (DDD) of a line are its orthogonal components along the North-South and East-West meridian directions, respectively. For a line of length lll and reduced bearing ╬╕\theta╬╕ in the SE quadrant (S ╬╕\theta╬╕ E), the latitude is directed South (lcosтБб╬╕l \cos \thetalcos╬╕) and the departure is directed East (lsinтБб╬╕l \sin \thetalsin╬╕).

ЁЯФв Key Formulas

Latitude┬а(L)=lcosтБб╬╕\text{Latitude } (L) = l \cos \thetaLatitude┬а(L)=lcos╬╕ тАФ Projection of line on the North-South axis

Departure┬а(D)=lsinтБб╬╕\text{Departure } (D) = l \sin \thetaDeparture┬а(D)=lsin╬╕ тАФ Projection of line on the East-West axis

тЪЩя╕П Working Principle

Latitude is the length of the projection of a line on the reference meridian (North-South line), calculated as L=lcosтБб╬╕L = l \cos \thetaL=lcos╬╕. Departure is the length of the projection on the perpendicular axis (East-West line), calculated as D=lsinтБб╬╕D = l \sin \thetaD=lsin╬╕. Since the bearing S50┬░E falls in the second quadrant (Southeast), the latitude is negative (South) and the departure is positive (East).

ЁЯУМ Key Points
  • тЦ╕

    Latitude is positive in North quadrant and negative in South quadrant.

  • тЦ╕

    Departure is positive in East quadrant and negative in West quadrant.

  • тЦ╕

    For bearing S50┬░E: Latitude is directed South (S) and Departure is directed East (E).

  • тЦ╕

    Calculation: L=50├ЧcosтБб(50┬░)=32.139┬аm┬а(S)L = 50 \times \cos(50┬░) = 32.139\text{ m (S)}L=50├Чcos(50┬░)=32.139┬аm┬а(S), D=50├ЧsinтБб(50┬░)=38.302┬аm┬а(E)D = 50 \times \sin(50┬░) = 38.302\text{ m (E)}D=50├Чsin(50┬░)=38.302┬аm┬а(E).

ЁЯЫая╕П Applications / Uses
  • тЦ╕

    Computation of consecutive coordinates in traverse surveying

  • тЦ╕

    Checking traverse closure error and adjusting closed traverses

ЁЯУД Additional Information
  • тЦ╕

    Option A gives North latitude and West departure, which corresponds to N50┬░W.

  • тЦ╕

    Option B gives North latitude and East departure, which corresponds to N50┬░E.

  • тЦ╕

    Option D gives South latitude and West departure, which corresponds to S50┬░W.

ЁЯУК Diagram / Illustration
NSWE50┬░l = 50mL = 50 cos(50┬░)= 32.13 m (S)D = 50 sin(50┬░)= 38.30 m (E)
тЬЕ

C is correct тАФ For a line bearing S50┬░E with length 50m, the latitude is 50cosтБб50┬░=32.13┬аm┬а(South)50 \cos 50┬░= 32.13\text{ m (South)}50cos50┬░=32.13┬аm┬а(South) and departure is 50sinтБб50┬░=38.30┬аm┬а(East)50 \sin 50┬░= 38.30\text{ m (East)}50sin50┬░=38.30┬аm┬а(East).

Core Concepts Used
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Latitude and Departure Reduced Bearing System Traverse Surveying
ЁЯТб EXAM TIP

Always identify the quadrant of the Reduced Bearing (RB) first to fix the directional signs (N/S for Latitude, E/W for Departure) before computing the trigonometric values.

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