\section*{Exercise 4 - V11013}

For a triangle, the following are given: \(a=5 \mathrm{~cm}, \beta=47^{\circ}\) and \(\gamma=55^{\circ}\).

Calculate \(b, c\) and \(\alpha\)!
}

Since the angles \(\beta\) and \(\gamma\) lie on the side \(a\), this triangle is uniquely constructible up to congruence according to the congruence theorems.

For the angle \(\alpha\), the interior angle sum theorem in the triangle gives

\[
\alpha=180^{\circ}-\beta-\gamma=180^{\circ}-47^{\circ}-55^{\circ}=78^{\circ}
\]

By the Law of Sines, we have

\[
\frac{b}{a}=\frac{\sin (\beta)}{\sin (\alpha)}
\]

and therefore

\[
b=a \cdot \frac{\sin (\beta)}{\sin (\alpha)}=5 \mathrm{~cm} \cdot \frac{\sin \left(47^{\circ}\right)}{\sin \left(78^{\circ}\right)} \approx 3.74 \mathrm{~cm}
\]

Similarly, by the Law of Sines,

\[
c=a \cdot \frac{\sin (\gamma)}{\sin (\alpha)}=5 \mathrm{~cm} \cdot \frac{\sin \left(55^{\circ}\right)}{\sin \left(78^{\circ}\right)} \approx 4.19 \mathrm{~cm}
\]