1979 AHSME Problems/Problem 24

Problem 24

Sides $AB,~ BC$, and $CD$ of (simple*) quadrilateral $ABCD$ have lengths $4,~ 5$, and $20$, respectively. If vertex angles $B$ and $C$ are obtuse and $\sin C = - \cos B =\frac{3}{5}$, then side $AD$ has length

  • A polygon is called “simple” if it is not self intersecting.

$\textbf{(A) }24\qquad \textbf{(B) }24.5\qquad \textbf{(C) }24.6\qquad \textbf{(D) }24.8\qquad \textbf{(E) }25$


We know that $\sin(C)=-\cos(B)=\frac{3}{5}$. Since $B$ and $C$ are obtuse, we have $\sin(180-C)=\cos(180-B)=\frac{3}{5}$. It is known that $\sin(x)=\cos(90-x)$, so $180-C=90-(180-C)=180-B$. We simplify this as follows:



Since $B+C=270^{\circ}$, we know that $A+D=360-(B+C)=90^{\circ}$. Now extend $AB$ and $CD$ as follows:

[asy] size(10cm); label("A",(-1,0)); dot((0,0)); label("B",(-1,4)); dot((0,4)); label("E",(-1,7)); dot((0,7)); label("C",(4,8)); dot((4,7)); label("D",(24,8)); dot((24,7)); draw((0,0)--(0,4)); draw((0,4)--(4,7)); draw((4,7)--(24,7)); draw((24,7)--(0,0)); draw((0,4)--(0,7), dashed); draw((0,7)--(4,7), dashed); //diagram by WannabeCharmander [/asy]

Let $AB$ and $CD$ intersect at $E$. We know that $\angle AED=90^{\circ}$ because $\angle E = 180 - (A+D)=180-90 = 90^{\circ}$.

Since $\sin BCD = \frac{3}{5}$, we get $\sin ECB=\sin(180-BCD)=\sin BCD = \frac{3}{5}$. Thus, $EB=3$ and $EC=4$ from simple sin application.

$AD$ is the hypotenuse of right $\triangle AED$, with leg lengths $AB+BE=7$ and $EC+CD=24$. Thus, $AD=\boxed{\textbf{(E)}25}$


See also

1979 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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