Determine the interior angles of the triangle $ABC$, given $b=2\,\mathrm{cm}$, $a=1.7\,\mathrm{cm}$ and $\alpha=30^\circ$. (We assume that the angles denoted by $\alpha$, $\beta$, $\gamma$ lie opposite of the sides $a$, $b$, $c$ consecutively.)
Student Peter's solution:
(1) Peter used the law of sines: $$\frac{\sin\beta}{b}=\frac{\sin\alpha}{a}$$
(2) He substituted the given values and got: $$\frac{\sin\beta}{2}=\frac{\sin 30^\circ}{1.7}$$
(3) Since $\sin 30^\circ=\frac12$, he obtained: $$\sin\beta=\frac{10}{17}$$
(4) Using a calculator, he found that equation $\sin\beta=\frac{10}{17}$ has a unique solution, namely: $$\beta=36^\circ$$
(5) Then, he calculated the remaining interior angle of the triangle: $$\alpha + \beta + \gamma = 180^\circ \implies \gamma=180^\circ - 30^\circ-36^\circ=114^\circ$$
(6) At last, he wrote all the interior angles of the triangle $ABC$: $$\alpha=30^\circ,\, \beta=36^\circ,\, \gamma=114^\circ$$
In one of the steps, Peter made a mistake. Find the mistake.
The mistake is in the step (4). The angle $\beta=36^\circ$ is not the only solution to the equation $\sin\beta=\frac{10}{17}$.
The mistake is in the step (1). The law of sines is not written correctly.
The mistake is in the step (3). Correctly, it should be that $\sin\beta= \frac{\sqrt3}{1{.}7}$.
The mistake is in the step (3). It holds $\sin\beta=\frac{10}{17}\cdot\frac14=\frac{5}{34}$.
The mistake is in the step (4). The angle $\beta=36^\circ$ is not the only solution to the equation $\sin\beta=\frac{10}{17}$. The obtuse angle $\beta=144^\circ$ also fits the specification. I.e., two triangles fit the assignment. One has interior angles $\alpha=30^\circ$, $\beta=36^\circ$, $\gamma=114^\circ$, the other one has interior angles $\alpha=30^\circ$, $\beta=144^\circ$, $\gamma=6^\circ$.