Problem 107

Let's break down the solution step by step, adhering to the rules:

For statement A:

Given $A < B$, we want to prove $\sin A < \sin B$.

- Since $A < B$ in a triangle, it implies that the side opposite to the smaller angle is shorter than the side opposite to the larger angle, i.e., $a < b$.
- By the Law of Sines, $\frac{a}{\sin A} = \frac{b}{\sin B}$.
- Since $a < b$, it follows that $\sin A < \sin B$.

Therefore, statement A is correct.

For statement B:

Given $a=2$ and $A=30^{\circ}$, we want to find the radius $R$ of the circumcircle.

- Using the Law of Sines, $R = \frac{a}{2\sin A} = \frac{2}{2\sin30^{\circ}} = \frac{2}{2 \times \frac{1}{2}} = \frac{2}{1} = 2$.

Hence, the radius of the circumcircle of $\triangle ABC$ is $2$, making statement B incorrect.

For statement C:

Given $\frac{a}{\cos A} = \frac{b}{\sin B}$, we aim to prove $A = 45^{\circ}$.

- By the Law of Sines, $\frac{a}{\sin A} = \frac{b}{\sin B}$.
- Equating $\frac{a}{\cos A} = \frac{a}{\sin A}$ implies $\cos A = \sin A$.
- The only angle for which $\cos A = \sin A$ in the range of $0^{\circ} < A < 180^{\circ}$ is $A = 45^{\circ}$.

Thus, statement C is correct.

For statement D:

Given $A=30^{\circ}$, $a=4$, and $b=3$, we need to determine the number of solutions for $\triangle ABC$.

- Using the Law of Cosines, $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.
- Substituting the given values, $\frac{\sqrt{3}}{2} = \frac{9 + c^2 - 16}{6c}$.
- Solving for $c$, we get $c = \frac{3\sqrt{3} \pm \sqrt{55}}{2}$.
- Since we only consider the positive root for the length of a side, $c$ has one positive solution.

Therefore, $\triangle ABC$ has one solution, making statement D incorrect.

The correct conclusions are $\boxed{\text{A and C}}$.