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Intersection, coordinates of B from S and A

Short answer

Base SA=304.019 m, bearing SA=134°38'34". In triangle SAB, ∠SBA=38°30'38", SB=404.034 m, AB=486.852 m. Bearing SB=48°59'45", so B = (1614.029 mE, 1435.592 mN).

The co-ordinates of stations S and A are (1309.12 m E, 1170.50 m N) and (1525.43 m E and 956.87 m N) respectively. Calculate the co-ordinates of point B which has been located by intersection from stations S and A observing the following angles. ∠BSA = 85°38'49" and ∠SAB = 55°50'33".

Coordinates are (Easting, Northing). B is taken on the left-hand (north-east) side of the line S to A; if it were on the other side, its position would be the mirror image of this one in the line SA.

           B
          / \
         /   \
        /85°  \ 55°
       S ------ A

Step 1: Base SA

ΔE=216.31,ΔN=−213.63\Delta E = 216.31,\quad \Delta N = -213.63 SA=216.312+(−213.63)2=304.019 m,bearing of SA=134°38′34"SA = \sqrt{216.31^2 + (-213.63)^2} = 304.019\ \text{m}, \qquad \text{bearing of SA} = 134°38'34"

Step 2: Triangle SAB

∠SBA=180∘−85∘38′49′′−55∘50′33′′=38°30′38"\angle SBA = 180^\circ - 85^\circ38'49'' - 55^\circ50'33'' = 38°30'38"

SB=SAsin⁡55∘50′33′′sin⁡38°30′38"=404.034 m,AB=SAsin⁡85∘38′49′′sin⁡38°30′38"=486.852 mSB = \frac{SA\sin 55^\circ50'33''}{\sin 38°30'38"} = 404.034\ \text{m}, \qquad AB = \frac{SA\sin 85^\circ38'49''}{\sin 38°30'38"} = 486.852\ \text{m}

Step 3: Coordinates of B

Bearing of SB = bearing of SA − 85°38'49" =48°59′45"= 48°59'45"

EB=1309.12+404.034sin⁡48°59′45"=1614.029 mE_B = 1309.12 + 404.034\sin 48°59'45" = 1614.029\ \text{m} NB=1170.50+404.034cos⁡48°59′45"=1435.592 mN_B = 1170.50 + 404.034\cos 48°59'45" = 1435.592\ \text{m}

Check from A: bearing of AB = bearing of AS + 55°50'33" =10°29′07"= 10°29'07", giving E=1614.029E = 1614.029 and N=1435.592N = 1435.592 (the same).

Answer: coordinates of B = (1614.029 E, 1435.592 N).

✓ Answer checked by Claude Opus 5.5 (high effort)

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