2. How many positive numbers are there among the first 100 terms of the sequence: $\sin 1^{\circ}, \sin 10^{\circ}, \sin 100^{\circ}, \sin 1000^{\circ}, \ldots ?$
2. Note that all members of the sequence, starting from $\sin 1000^{\circ}$, are equal to each other, since the difference between the numbers $10^{k+1}$ and $10^{k}$ for natural $k>2$ is a multiple of 360. Indeed, $10^{k+1}-10^{k}=10^{k}(10-1)=9 \cdot 10^{k}=$ $9 \cdot 4 \cdot 10 \cdot 25 \cdot 10^{k-3} \vdots 360$. By direct verification, we can see that $\sin 1^{\circ}>0$, $\sin 10^{\circ}>0, \sin 100^{\circ}>0, \sin 1000^{\circ}=\sin \left(360^{\circ} \cdot 3-80^{\circ}\right)=\sin \left(-80^{\circ}\right)<0$. Therefore, the rest of the members of the sequence will also be negative. Thus, among the first 100 members of the sequence, there will be exactly three positive members.

Answer: 3.