Area of the triangle II

Determine the area of triangle $ABC$. The lengths of two sides are given: $$|BC|=a=3\,\mathrm{cm}\ \mbox{ and }\ |AB|=c=2\,\mathrm{cm}$$ Also, the measures of two interior angles are given:

$$|\sphericalangle BAC|=\alpha= 44^\circ \ \mbox{ and }|\sphericalangle ABC| = \beta = 108.4^\circ$$ Round the result (in $\mathrm{cm}^2$) to one decimal place.

Josef used the given side lengths and angles to calculate the area of the triangle using the formula: $$S=\frac12\cdot a\cdot c\cdot \sin\beta$$ His calculation was: $$S = \frac12\cdot 3\cdot 2\cdot \sin108.40^\circ \cong 2.8\,\mathrm{cm}^2$$ Tonda took a longer approach and calculated in the following steps:

1. He calculated the missing interior angle: $$\gamma=180^\circ-(\alpha+\beta)=180^\circ -(44^\circ+108.4^\circ)=27.6^\circ$$
2. Using the Law of Sines, he derived how to calculate the length of side $|AC| = b$: $$\frac{b}{\sin\beta}=\frac{a}{\sin\alpha}\Rightarrow b=\frac{\sin\beta}{\sin\alpha}\cdot a$$
3. He then calculated the area of the triangle using the formula: $$S=\frac12\cdot b\cdot c\cdot \sin\gamma=\frac12\cdot \frac{\sin\beta}{\sin\alpha}\cdot a\cdot c\cdot \sin\gamma$$

Substituting the values he found: $$S=\frac12\cdot\frac{\sin108.4^\circ}{\sin44^\circ}\cdot 3\cdot 2\cdot \sin27.6^\circ\approx 1.9\,\mathrm{cm}^2$$ It is clear that at least one student made a mistake in their calculation. Who made the mistake, and in which step?