7.38 From the right angle vertex of a right triangle, draw lines to the trisection points of the hypotenuse. These two line segments are of lengths $\sin x$ and $\cos x$, where $x$ is a real number such that $0 < x < \frac{\pi}{2}$. The length of the hypotenuse is 
(A) $\frac{4}{3}$.
(B) $\frac{3}{2}$.
(C) $\frac{3 \sqrt{5}}{5}$.
(D) $\frac{2 \sqrt{5}}{3}$.
(E) There is no unique solution according to the given conditions.
(31st American High School Mathematics Examination, 1980)
[Solution] As shown in the figure, applying the Pythagorean theorem in $\triangle C D F$ and $\triangle C E G$, we get
$$
\left\{\begin{array}{l}
4 a^{2}+b^{2}=\sin ^{2} x \\
a^{2}+4 b^{2}=\cos ^{2} x .
\end{array}\right.
$$
Adding (1) and (2) gives $\quad 5\left(a^{2}+b^{2}\right)=1$.

Therefore, $A B=3 \sqrt{a^{2}+b^{2}}=3 \sqrt{\frac{1}{5}}$
$$
=\frac{3 \sqrt{5}}{5} \text {. }
$$

Thus, the correct choice is $(C)$.