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
\( \angle DEC = 1'' \)
\( \angle DEF = 170^{\circ} 20' \)
Show 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.
This is a shaft correlation problem solved with the Weisbach triangle idea: two plumb wires C and D hang close together, an underground station E sights both of them and also sights the next survey point F. Because C and D are so close and E is comparatively far off, triangle CDE is almost a straight line, and that lets us carry the surface bearing of CD down to the underground bearing of EF.
Bearing of CD from coordinates: with C at (375, 1120) and D at (376, 1121), the easting and northing both increase by 1 m, so the line runs exactly northeast, at a whole circle bearing of \( 45^{\circ} \).
Angle at the middle point, C: in triangle CDE, angle D and angle E are both tiny, so angle C is almost a straight \( 180^{\circ} \). Since the three angles add to \( 180^{\circ} \), the shortfall from a straight line at C equals the sum of the other two, \( 180^{\circ} - \angle C = \angle D + \angle E \). Using the sine rule as before, \( \angle D = 1.4'' \), and \( \angle E = 1'' \) is given, so the deviation at C from a straight line is \( 1.4'' + 1'' = 2.4'' \).
Bearing of CE: since D, C, E run almost straight, the direction C to E continues almost the same as D to C, \( 225^{\circ} \), tipped by the \( 2.4'' \) deviation at C, giving \( \text{bearing}(C \to E) = 225^{\circ}00'02.4'' \). Reversing gives \( \text{bearing}(E \to C) = 45^{\circ}00'02.4'' \).
Angle between EC and ED at station E: the measured angle \( \angle DEC = 1'' \) is exactly the gap between bearing(E to D) and bearing(E to C): \( 45^{\circ}00'02.4'' - 45^{\circ}00'01.4'' = 1'' \). This checks out, confirming the working is consistent.
Angle from EC to EF: since EC sits \( 1'' \) further round than ED, the angle from EC to EF is \( 1'' \) less than the angle from ED to EF, \( \angle CEF = 170^{\circ}20' - 1'' = 170^{\circ}19'59'' \).
Coming at it through C instead of through D lands on exactly the same figure, \( 215^{\circ}20'1.4'' \), which confirms the first method was not a fluke.
Let's summarize:
The coordinates give the surface bearing of CD as \( 45^{\circ} \) directly.
The Weisbach triangle's tiny angles, 1.4 seconds at D and 2.4 seconds at C, carry that bearing down to station E with almost no change.
Adding the measured angle DEF, or the equivalent angle CEF, at E gives the bearing of EF as \( 215^{\circ}20'1.4'' \).
Note for the student: the official key lists this question as MTA (marks given to all), most likely because the direction of the tiny correction angle depends on a left or right reading of the figure that is not fully settled on paper. This working supports option (C).
