### Instruction: In triangle $ABC,$ $\sin A = \frac{3}{5}$ and $\sin B = \frac{24}{25}.$  Enter all the possible values of $\sin C,$ separated by commas.### Response: To solve this problem, I need to use the fact that the sum of the angles in a triangle is $180^\circ.$
Let me start by finding the values of $A$ and $B$ in degrees.
I can use a calculator or the inverse sine function to do this.
For $A,$ I get that $\sin^{-1} \frac{3}{5} \approx 36.87^\circ$ or $180^\circ - 36.87^\circ \approx 143.13^\circ.$
For $B,$ I get that $\sin^{-1} \frac{24}{25} \approx 73.74^\circ$ or $180^\circ - 73.74^\circ \approx 106.26^\circ.$
Now, I need to check which combinations of these values are possible for $A$ and $B$ in a triangle.
I can use the fact that the largest angle in a triangle must be smaller than the sum of the other two angles.
If $A = 36.87^\circ$ and $B = 73.74^\circ,$ then $C = 180^\circ - (36.87^\circ + 73.74^\circ) = 69.39^\circ.$
This is a valid triangle, so I can find $\sin C = \sin 69.39^\circ \approx 0.93.$
If $A = 36.87^\circ$ and $B = 106.26^\circ,$ then $C = 180^\circ - (36.87^\circ + 106.26^\circ) = 36.87^\circ.$
This is also a valid triangle, and I can find $\sin C = \sin 36.87^\circ = \frac{3}{5}.$
If $A = 143.13^\circ$ and $B = 73.74^\circ,$ then $C = 180^\circ - (143.13^\circ + 73.74^\circ) = -36.87^\circ.$
This is not a valid triangle, because the angle $C$ is negative.
If $A = 143.13^\circ$ and $B = 106.26^\circ,$ then $C = 180^\circ - (143.13^\circ + 106.26^\circ) = -69.39^\circ.$
This is also not a valid triangle, for the same reason.
Therefore, the only possible values of $\sin C$ are $0.93$ and $\frac{3}{5}.$