The sides of a triangle $ABC$ are: $|AB|=5$, $|BC|=5\sqrt{7}$, and $|AC|=15$. The measure of $\measuredangle BAC$ is $60^{\circ}$. Two girls, Judith and Margaret, calculated the measure of $\measuredangle ABC$ but got different results.
Judith used the Law of Cosines: $$ \begin{gather} |AC|^2=|AB|^2+|BC|^2−2|AB||BC|\cos (\measuredangle ABC) \cr 15^2=5^2+(5\sqrt{7})^2−2\cdot 5\cdot 5\sqrt{7}\cdot \cos (\measuredangle ABC) \cr 225=25+175−50\sqrt{7}\cdot\cos (\measuredangle ABC) \cr \cos (\measuredangle ABC)=−\frac{25}{50\sqrt{7}} \cr \measuredangle ABC\approx 101^{\circ} \end{gather} $$
Margaret used the Law of Sines: $$ \begin{gather} \frac{|BC|}{\sin(\measuredangle BAC)}=\frac{|AC|}{\sin(\measuredangle ABC) }\cr \frac{5\sqrt{7}}{\frac{\sqrt{3}}{2}}=\frac{15}{\sin(\measuredangle ABC)}\cr \sin(\measuredangle ABC)=\frac{3\sqrt{3}}{2\sqrt{7}} \cr \measuredangle ABC\approx 79^{\circ} \end{gather} $$ Here are some comments. Which one is correct?
Judith demonstrated the right solution, $\measuredangle ABC\approx 101^{\circ}$. If the lengths of three sides of a triangle are given, The Law of Cosines can be used to solve the triangle.
Both results are correct. There may be two possible measures for $\measuredangle ABC$: $\measuredangle ABC\approx 101^{\circ}$ or $\measuredangle ABC\approx 79^{\circ}$.
Margaret showed the right solution, $\measuredangle ABC\approx 79^{\circ}$. An obtuse angle of $101^{\circ}$ cannot be the right solution. Since the sine of an obtuse angle is negative, the area of the triangle would be negative, which is not possible. See the formula for the area of $\Delta ABC$: $$ A=|AB||BC|\sin(\measuredangle ABC) $$
The example can be solved using both the Law of Sines and the Law of Cosines. Unfortunately, both girls made a mistake and their results are wrong.
If we use the Law of Sines, we get two measures for $\measuredangle ABC$: $\sin(\measuredangle ABC)=\frac{3\sqrt{3}}{2\sqrt{7}}$, hence $\measuredangle ABC\approx 79^{\circ}$ or $\measuredangle ABC=180^{\circ}−79^{\circ}\approx 101^{\circ}$ ($\sin \alpha=\sin(180^{\circ}−\alpha$)). Calculating the measure of $\measuredangle ACB$ will eliminate one of these results: $$ \begin{gather} \frac{|BC|}{\sin(\measuredangle BAC)}=\frac{|AB|}{\sin(\measuredangle ACB )}\cr \sin(\measuredangle ACB )=\frac{\sqrt{3}}{2\sqrt{7}} \cr \measuredangle ACB\approx 19^{\circ} \lor \measuredangle ACB\approx 161^{\circ} \end{gather} $$ Since $\measuredangle BAC=60^{\circ}$ and $\measuredangle ACB+\measuredangle ABC+\measuredangle BAC=180^{\circ}$, the measure of $\measuredangle ABC$ must be about $101^{\circ}$.