Question

The observations from a correlation survey are shown. If the coordinates of points C and D are (East: 375 m, North: 1120 m) and (East: 376 m, North: 1121 m), respectively, the whole circle bearing of the line EF is

CD = 3.0 m
CE = 4.2 m
\( \angle DEC = 1'' \)
\( \angle DEF = 170^{\circ} 20' \)

• A. \( 215^{\circ}\,19'\,59.6'' \)
• B. \( 35^{\circ}\,19'\,59.6'' \)
• C. \( 215^{\circ}\,20'\,1.4'' \)
• D. \( 35^{\circ}\,20'\,1.4'' \)

Hint:

Find the bearing of CD from the coordinates first, then use the sine rule on the tiny angle DEC to carry that bearing down to station E before adding angle DEF.

Answer 1

The Correct Option is C

Step 1: Get the bearing of CD from the coordinates.
The whole circle bearing of a line is measured clockwise from north, and for a line from point 1 to point 2 it is found from \( \theta = \tan^{-1}\left(\dfrac{\Delta E}{\Delta N}\right) \), taking the correct quadrant.
\( \Delta E = 376 - 375 = 1 \) m, \( \Delta N = 1121 - 1120 = 1 \) m, both positive, so line CD lies in the northeast quadrant.
\[ \theta_{CD} = \tan^{-1}\left(\frac{1}{1}\right) = 45^{\circ} \]
So the bearing of C to D is \( 45^{\circ} \), and the reverse bearing, D to C, is \( 45^{\circ} + 180^{\circ} = 225^{\circ} \).

Step 2: Solve the Weisbach (near-collinear) triangle CDE for the small angle at D.
Points D, C and E lie almost in a straight line down the shaft; the angle at E between the two wires, \( \angle DEC = 1'' \), is tiny because C and D are close together while E is well below. By the sine rule in triangle CDE, side CD is opposite the angle at E, and side CE is opposite the angle at D:
\[ \frac{CD}{\sin(\angle E)} = \frac{CE}{\sin(\angle D)} \quad\Rightarrow\quad \sin(\angle D) = \frac{CE}{CD}\sin(\angle E) \]
Because \( 1'' \) is such a tiny angle, \( \sin(\angle E) \approx \angle E \) in radians, and the same holds for \( \angle D \), so
\[ \angle D \approx \frac{CE}{CD}\times \angle E = \frac{4.2}{3.0}\times 1'' = 1.4'' \]

Step 3: Carry this small angle onto the bearing of DE, then reverse it to get the bearing of ED.
The direction D to E deviates from the direction D to C by this small angle \( \angle D \), so
\[ \text{bearing}(D \to E) = 225^{\circ} + 1.4'' = 225^{\circ}00'01.4'' \]
Reversing this line gives the bearing E to D:
\[ \text{bearing}(E \to D) = 225^{\circ}00'01.4'' - 180^{\circ} = 45^{\circ}00'01.4'' \]

Step 4: Add the measured angle DEF to reach the bearing of EF.
The angle \( \angle DEF = 170^{\circ}20' \) is measured clockwise from line ED to line EF at station E, so
\[ \text{bearing}(E \to F) = 45^{\circ}00'01.4'' + 170^{\circ}20'00'' = 215^{\circ}20'01.4'' \]

Final Answer:
Working the coordinates and the Weisbach triangle through cleanly gives \( 215^{\circ}20'1.4'' \), option (C). Note for the student: the official key for this question was marked MTA (marks awarded to all candidates), most likely because the sign of the tiny correction angle at D (which side the wires deviate to, as seen from station E) cannot be pinned down from a flat page diagram alone, and reading it the other way lands about 2.8 seconds away from this value, close to but not exactly matching option (A). This working supports option (C). \[ \boxed{215^{\circ}20'1.4''} \]