Answer
The student's statement is true.
We can prove it by setting up a random case in which 2 angles, for example $A$ and $C$, and a non-included side are known; another case in which 2 angles and their included side are known.
Then we show that in both cases, we can always figure out the rest of the angles and sides of the triangle.
That means we can solve the triangle given the known variables, and the triangle is uniquely determined.
Work Step by Step
The law of sines for a triangle with side $a,b,c$ and angle $A, B, C$ is
$$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$$
The statement of the student is "If we know any two angles and one side of a triangle, then the triangle is uniquely determined.'
Being uniquely determined means that there is only one possibility of the triangle with these known angles and side. In any case, that means we can solve a triangle if we are given any two angles and one side of it.
To prove this is true, I would suppose a random case in which angles $A$ and $C$ and a non-included side are known, and another case in which angles $A$ and $C$ and included side $b$ are known. I would prove that using the law of sines, we can always solve the triangle, meaning that the triangle is uniquely determined.
1) Angles $A$ and $C$ are known, non-included side $a$ is known
- First, we can immediately calculate $c$ by applying the law of sines:
$$\frac{a}{\sin A}=\frac{c}{\sin C}$$
$$c=\frac{a\sin C}{\sin A}$$
Since $a, A, C$ are known, find $c$ is possible.
- Next, angle $B$ can be calculated as we apply the sum of 3 angles in a triangle equation.
$$A+B+C=180^\circ$$
$$B=180^\circ-A-C$$
- Finally, side $b$ can be calculated, again, with the law of sines, since angle $B$ is now known.
$$\frac{a}{\sin A}=\frac{b}{\sin B}$$
$$b=\frac{a\sin B}{\sin A}$$
Thus, the triangle has been solved. Generalizing for all cases, we conclude that given any two angles and a non-included side of a triangle, the triangle is uniquely determined.
2) Angles $A$ and $C$ are known, included side $b$ is known
- First, angle $B$ can be calculated as we apply the sum of 3 angles in a triangle equation.
$$A+B+C=180^\circ$$
$$B=180^\circ-A-C$$
- Now that both $b$ and $B$ are known, meaning that $\frac{b}{\sin B}$ is known, and also angles $A$ and $C$ are known, we can apply the law of sines to find $a$ and $c$.
$$\frac{a}{\sin A}=\frac{b}{\sin B}$$
$$a=\frac{b\sin A}{\sin B}$$
$$\frac{c}{\sin C}=\frac{b}{\sin B}$$
$$c=\frac{b\sin C}{\sin B}$$
Thus, the triangle has been solved. Generalizing for all cases, we conclude that given any two angles and an included side of a triangle, the triangle is also uniquely determined.
Therefore, we can conclude that if we know any two angles and one side of a triangle, then the triangle is uniquely determined, and it is possible to solve the triangle. The student's statement is true as a result.
