\section*{Exercise 1 - 011011}

For an isosceles triangle, the following are given: \(A B=c=87.51 \text{ m}, \angle C A B=\alpha=93.42^{\circ}\). Calculate the remaining angles and sides!
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![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1175.jpg?height=317&width=325&top_left_y=590&top_left_x=337)

Since \(2 \alpha > 180^{\circ}\), \(\alpha\) cannot be a base angle. For the size of the two base angles \(\angle A B C = \beta\) and \(\angle A C B = \gamma\), we have

\[
\beta = \gamma = \frac{1}{2} \left(180^{\circ} - \alpha\right) = 43.29^{\circ}
\]

Since \(A C = b\) and \(A B\) are legs, it follows that \(b = c = 87.51 \, \text{m}\). If we draw the height \(A D\) perpendicular to \(B C = a\), we obtain in the resulting right triangle \(A B D\) the equation \(\cos \beta = \frac{a}{2 c}\) and thus \(a = 2 c \cdot \cos \beta = 127.40 \, \text{m}\).