$17 \cdot 78$ Two rays starting from $O$ form an angle of $30^{\circ}$, point $A$ is on one ray, point $B$ is on the other, and $A B=1$. Then the maximum length of $O B$ is
(A) 1 .
(B) $\frac{1+\sqrt{3}}{\sqrt{2}}$
(C) $\sqrt{3}$.
(D) 2 .
(E) $\frac{4}{\sqrt{3}}$.
(46th American High School Mathematics Examination, 1995)
[Solution] Let $O A=x, O B=y$. In $\triangle A B O$, by the cosine rule we have
$$
x^{2}+y^{2}-\sqrt{3} x y=1 \text {, }
$$

which simplifies to $x^{2}-\sqrt{3} y x+\left(y^{2}-1\right)=0$.
Considering the above equation as a quadratic in $x$ with real roots, we have
$$
\Delta=3 y^{2}-4\left(y^{2}-1\right) \geqslant 0 \text {. }
$$

This gives $y \leqslant 2$, so $y_{\text {max }}=2$.
Therefore, the answer is $(D)$.