Does there exist an obtuse triangle $ABC$ with $b=3\, \mathrm{cm}$, $a=4\, \mathrm{cm}$, and $\alpha=55^\circ$? (We assume the angles $\alpha$, $\beta$, $\gamma$ lie opposite the sides $a$, $b$, $c$, respectively.) If so, find its interior angle $\beta$.
Martina's solution:
(1) According to the law of sines: $$\frac{b}{\sin\beta}=\frac{a}{\sin\alpha}$$
(2) After rearranging, we get: $$\sin\beta=\frac{b}{a}\cdot\sin\alpha$$
(3) Substituting the known values (using a calculator) we obtain: $$\sin\beta\cong 0.61436$$
(4) Since $\sin\beta>0$, the solution to the equation is an acute angle $\beta\cong37.9^\circ$ and an obtuse angle $\beta\cong142.1^\circ$.
(5) The angle $\beta=37.9^\circ$ does not meet the problem's conditions. Triangle $ABC$ would not be obtuse.
(6) There is only a single obtuse triangle $ABC$ with the given properties. Its internal angle $\beta$ is approx. $142.1^\circ$.
Martina made a mistake in one of her reasonings. Find the mistake that Martina made.
The mistake is in step (6). Even the angle $\beta\cong142.1^\circ$ does not match the specification. A triangle with the given properties doesn't exist.
The mistake is in step (1). The law of sines is not written correctly.
The mistake is in step (2). Expression of $\sin\beta$ from the law of sines is incorrect.
The mistake is in step (3). We get $\sin\beta=-0.7498$ after substituting. This means that a triangle with the given properties does not exist.
The mistake is in step (5). Although the angle $\beta$ is acute, the obtuse angle in the triangle will be the angle $\gamma$.
The sum of the interior angles of a triangle is always equal to $180^\circ$, i.e. $$\alpha+\beta+\gamma = 180^\circ$$
The sum of any two angles cannot be greater than or equal to $180^\circ$, because then the third angle would have to be zero or negative, which is not possible in a triangle.
In our case, we found that
$$\alpha+\beta > 180^\circ$$
and therefore such a triangle does not exist.