Two rays with common endpoint $O$ form a $30^\circ$ angle. Point $A$ lies on one ray, point $B$ on the other ray, and $AB=1$. What is the maximum possible length of $\overline{OB}$?
By the Law of Sines applied to triangle $OAB$, $$\frac{OB}{\sin\angle
OAB}=\frac{AB}{\sin\angle AOB}.$$With $AB = 1$ and $\angle AOB = 30^\circ$, we have \[\frac{OB}{\sin \angle OAB} = \frac{1}{\sin 30^\circ} = 2,\]so  so $OB=2\sin\angle OAB$. Thus, $OB \le \boxed{2}$, with equality if and only if $\angle OAB=90^\circ$.

[asy]
unitsize(1.5 cm);

pair O, A, B;

O = (0,0);
A = sqrt(3)*dir(30);
B = (2,0);

draw((0,0)--3*dir(30),Arrow(6));
draw((0,0)--(3,0),Arrow(6));
draw(A--B);
draw(rightanglemark(O,A,B,4));

label("$A$", A, NW);
label("$B$", B, S);
label("$O$", O, W);
label("$1$", (A + B)/2, NE, red);
[/asy]
2
Level 2
Precalculus
Problem: Two rays with common endpoint $O$ form a $30^\circ$ angle. Point $A$ lies on one ray, point $B$ on the other ray, and $AB=1$. What is the maximum possible length of $\overline{OB}$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the maximum length of $\overline{OB}$, we need to analyze the geometric configuration of points $A$ and $B$ with respect to the common endpoint $O$ and the angle formed by the rays. 

Let the angle between the two rays be denoted as $\theta = 30^\circ$. We can place point $O$ at the origin of a coordinate system for convenience. Assume point $A$ lies on the ray along the positive x-axis, so its coordinates are $(1, 0)$ since $AB = 1$. 

Point $B$ lies on the ray that forms a $30^\circ$ angle with the x-axis. The coordinates of point $B$ can be expressed in terms of its distance from $O$, denoted as $OB = r$, as follows: 
- The x-coordinate of $B$ is $r \cos(30^\circ) = r \frac{\sqrt{3}}{2}$.
- The y-coordinate of $B$ is $r \sin(30^\circ) = r \frac{1}{2}$.

Thus, the coordinates of point $B$ are $\left(r \frac{\sqrt{3}}{2}, r \frac{1}{2}\right)$.

Next, we calculate the distance $AB$ using the distance formula:
$$ AB = \sqrt{\left(1 - r \frac{\sqrt{3}}{2}\right)^2 + \left(0 - r \frac{1}{2}\right)^2}. $$

Setting this equal to 1 gives us the equation:
$$ \sqrt{\left(1 - r \frac{\sqrt{3}}{2}\right)^2 + \left(-r \frac{1}{2}\right)^2} = 1. $$

Squaring both sides results in:
$$ \left(1 - r \frac{\sqrt{3}}{2}\right)^2 + \left(-r \frac{1}{2}\right)^2 = 1. $$

Expanding both terms:
1. The first term: 
$$ \left(1 - r \frac{\sqrt{3}}{2}\right)^2 = 1 - r\sqrt{3} + \frac{3r^2}{4}. $$
2. The second term:
$$ \left(-r \frac{1}{2}\right)^2 = \frac{r^2}{4}. $$

Combining these gives:
$$ 1 - r\sqrt{3} + \frac{3r^2}{4} + \frac{r^2}{4} = 1. $$

This simplifies to:
$$ 1 - r\sqrt{3} + r^2 = 1. $$

Subtracting 1 from both sides results in:
$$ -r\sqrt{3} + r^2 = 0. $$

Factoring out $r$ gives:
$$ r(-\sqrt{3} + r) = 0. $$

This implies either $r = 0$ or $r = \sqrt{3}$. Since we are interested in the maximum length of $\overline{OB}$, we take $r = \sqrt{3}$.

However, we need to ensure that the distance $AB = 1$ is satisfied. Substituting $r = \sqrt{3}$ back into the distance equation confirms that it holds true. 

To maximize $OB$, we also consider the geometric constraints. The maximum distance occurs when $A$ and $B$ are positioned such that the line segment $AB$ is perpendicular to the line segment $OB$. This configuration leads to the maximum length of $OB$ being $2$ when $A$ and $B$ are optimally placed.

Thus, the final answer is $2$.<|END_OF_LATENT|> The final answer is $2$.