$7 \cdot 40$ A simple (i.e., non-intersecting) quadrilateral $A B C D$ has sides $A B, B C$, and $C D$ of lengths 4, 5, and 20, respectively. If angles $B$ and $C$ are obtuse, and $\sin C=-\cos B=$ $\frac{3}{5}$, then the length of side $A D$ is
(A) 24 .
(B) 24.5 .
(C) 24.6 .
(D) 24.8 .
(E) 25 .
(30th American High School Mathematics Examination, 1979)
［Solution］Let the intersection point of the extensions of line segments $AB$ and $DC$ be $E$, and let $\beta$ and $\theta$ represent $\angle EBC$ and $\angle ECB$, respectively. Since
$$
\cos \beta = -\cos B = \sin C = \sin \theta,
$$

and both $\beta$ and $\theta$ are acute angles, it follows that $\beta + \theta = 90^{\circ}$.
Therefore, $\angle BEC$ is a right angle.

Also, since $\quad BE = BC \sin \theta = 3, \quad CE = BC \sin \beta = 4$,
we have $AE = 7, DE = 24$, and $AD$ is the hypotenuse of the right triangle $\triangle ADE$, so
$$
AD = \sqrt{24^2 + 7^2} = 25.
$$

Thus, the answer is $(E)$.