Question

It is needed to establish the coordinates of a station A. Two stations B and C are available whose coordinates are given in the table.
Length of line CA = 1212.38 m; \(\angle BCA = 64^\circ\).
The Easting of station A is ______ m (Rounded off to two decimal places).
Assume station A lies to the west of the traverse line BC.

Station	Easting (m)	Northing (m)
B	11054.09	9484.37
C	10827.62	10112.15

Hint:

Find the bearing of BC from the coordinates, add or subtract the given angle at C to get the bearing of CA, then resolve the length CA along that bearing, choosing the direction that places A west of line BC.

Answer 1

Correct Answer: 9980

Step 1: Compute the whole circle bearing (azimuth) of line BC.
Using the coordinate difference method, the departure and latitude from B to C are: \[ \Delta E = E_C - E_B = 10827.62 - 11054.09 = -226.47 \text{ m}, \quad \Delta N = N_C - N_B = 10112.15 - 9484.37 = 627.78 \text{ m} \] Since \(\Delta E\) is negative and \(\Delta N\) is positive, line BC lies in the north-west quadrant. The reference angle from north is \[ \theta = \tan^{-1}\left(\frac{226.47}{627.78}\right) = \tan^{-1}(0.36075) \approx 19^\circ50' \] so the azimuth of BC (from B) is \[ \text{Az}_{BC} = 360^\circ - 19^\circ50' = 340^\circ10' \] and the back azimuth CB is \[ \text{Az}_{CB} = 340^\circ10' - 180^\circ = 160^\circ10' \]
Step 2: Apply the angle BCA at station C to get the azimuth of CA.
\(\angle BCA = 64^\circ\) is measured at C between ray CB and ray CA, giving two possible azimuths for CA: \(160^\circ10' + 64^\circ = 224^\circ10'\) or \(160^\circ10' - 64^\circ = 96^\circ10'\). Since A must lie to the WEST of line BC, only the first option carries A to that side (it has a south-westerly, negative-easting departure from C); the second option, \(96^\circ10'\), moves east of the line and is rejected. So \[ \text{Az}_{CA} = 224^\circ10' \]
Step 3: Compute the departure and latitude of CA.
With \(CA = 1212.38\) m: \[ \Delta E_{CA} = 1212.38 \times \sin(224^\circ10') = 1212.38 \times (-0.6967) = -844.67 \text{ m} \] \[ \Delta N_{CA} = 1212.38 \times \cos(224^\circ10') = 1212.38 \times (-0.7173) = -869.68 \text{ m} \]
Step 4: Compute the Easting of A.
\[ E_A = E_C + \Delta E_{CA} = 10827.62 - 844.67 = 9982.95 \text{ m} \]
Step 5: State the result.
\[ \boxed{E_A \approx 9982.95 \text{ m}} \]