Distances OA and OC by resection on A, B, C

The sides AB = 1200 m, BC = 1442 m and CA = 1960 m are used as given (the value marked [?] in the question is taken as printed). The angles $\alpha = \angle AOB = 45^\circ30'$ and $\beta = \angle BOC = 52^\circ15'$ are observed at O. B is the middle station. The method is the tangent method (Pothenot's problem).

Step 1: Angle B of triangle ABC

$$\cos B = \frac{AB^2 + BC^2 - CA^2}{2\,AB\cdot BC} = \frac{1200^2 + 1442^2 - 1960^2}{2(1200)(1442)} = -0.09311$$

$B = 95°20'33"$

Step 2: Sum and ratio of the unknown angles

Let $x = \angle OAB$ and $y = \angle OCB$. In the quadrilateral OABC:

$$x + y = S = 360^\circ - (\alpha + \beta + B) = 360^\circ - (45^\circ30' + 52^\circ15' + 95°20'33") = 166°54'27"$$ $$k = \frac{\sin x}{\sin y} = \frac{BC\sin\alpha}{AB\sin\beta} = \frac{1442\sin 45^\circ30'}{1200\sin 52^\circ15'} = 1.08398$$

Step 3: Solve for x and y

$$\tan\frac{x-y}{2} = \frac{k-1}{k+1}\tan\frac{S}{2} = \frac{0.08398}{2.08398}\tan 83°27'13" \quad\Rightarrow\quad \frac{x-y}{2} = 19°20'56"$$ $$x = \frac{S}{2} + 19°20'56" = 102°48'10",\qquad y = \frac{S}{2} - 19°20'56" = 64°06'17"$$

Step 4: Distances OA and OC

In triangle AOB: $\angle ABO = 180^\circ - \alpha - x = 31°41'50"$

$$OA = \frac{AB\sin\angle ABO}{\sin\alpha} = \frac{1200\sin 31°41'50"}{\sin 45^\circ30'} = 884.005\ \text{m}$$

In triangle BOC: $\angle CBO = 180^\circ - \beta - y = 63°38'43"$

$$OC = \frac{BC\sin\angle CBO}{\sin\beta} = \frac{1442\sin 63°38'43"}{\sin 52^\circ15'} = 1634.172\ \text{m}$$

Check: OB from triangle AOB $= \frac{1200\sin x}{\sin\alpha} = 1640.611$ m and from triangle BOC $= \frac{1442\sin y}{\sin\beta} = 1640.611$ m. They agree.

Answer: OA = 884.01 m and OC = 1634.17 m (OB = 1640.61 m).