$17 \cdot 81$ A certain triangle has a base of 80, one base angle of $60^{\circ}$, and the sum of the other two sides is 90. Then the shortest side is
(A) 45.
(B) 40.
(C) 36.
(D) 17.
(E) 12.
(10th American High School Mathematics Examination, 1959)
[Solution] In $\triangle ABC$, if $AB=80, BC=a$, then $AC=b=90-a$, and $\angle B=60^{\circ}$.
Draw the altitude $CD$ on $AB$, and let $BD=x$, then
$$
CD=\sqrt{3} x, \quad a=2 x, \quad b=90-2 x .
$$

By the Pythagorean theorem, we have $3 x^{2}+(80-x)^{2}=(90-2 x)^{2}$, which gives $x=\frac{17}{2}$, and $a=17, b=73$. Therefore, the answer is $(D)$.