\section*{Exercise 2 - V11022}

On the left bank of a river, whose banks are straight and parallel, a baseline \(A B=250 \mathrm{~m}\) has been laid out (measurement error \(\pm 0.50 \mathrm{~m}\)). The point \(C\) on the right bank is sighted, and the angles \(\measuredangle C A B=41^{\circ}\) and \(\measuredangle A B C=72^{\circ}\) are measured.

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1169.jpg?height=60&width=1314&top_left_y=1740&top_left_x=268)

a) Calculate the width \(x\) of the river without considering the measurement errors!

b) Calculate the possible error of the result.
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![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1169.jpg?height=386&width=636&top_left_y=1937&top_left_x=710)

a) General solution: We have \(\sin \beta = \frac{C D}{C B} = \frac{h_{c}}{a}\) and \(h_{c} = a \sin \beta\). From the Law of Sines, by substitution,

\[
\frac{a}{\sin \alpha} = \frac{c}{\sin \gamma} \Rightarrow a = \frac{c \sin \alpha}{\sin \gamma} \quad \Rightarrow \quad h_{c} = \frac{c \cdot \sin \alpha \cdot \sin \beta}{\sin \left(180^{\circ} - \alpha - \beta\right)}
\]

For the given values, this results in \(h_{c} = 169.5 \, \text{m}\). The width of the river, without considering measurement errors, is \(169.5 \, \text{m}\).

b) If the height \(h_{c}\) is calculated with \(c + 0.5\), \(\alpha + 0.5^{\circ}\), and \(\beta + 0.5^{\circ}\), it results in \(h_{c} = 173.3 \, \text{m}\).

In the case of \(c - 0.5\), \(\alpha - 0.5^{\circ}\), and \(\beta - 0.5^{\circ}\), \(h_{c} = 165.7 \, \text{m}\).

The possible error is thus \(\frac{173.5 - 165.5}{2} = 4 \, \text{m}\).