Difference between revisions of "1979 AHSME Problems/Problem 24"

Latest revision as of 23:23, 21 June 2018

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.

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

Solution

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:

$-90+C=180-B$

$B+C=270^{\circ}$

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, $AD=\boxed{\textbf{(E)}25}$

-WannabeCharmander

1979 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

Difference between revisions of "1979 AHSME Problems/Problem 24"

Latest revision as of 23:23, 21 June 2018

Problem 24

Solution

See also

@@ Line 12: / Line 12: @@
 ==Solution==
+We know that <math>\sin(C)=-\cos(B)=\frac{3}{5}</math>. Since <math>B</math> and <math>C</math> are obtuse, we have <math>\sin(180-C)=\cos(180-B)=\frac{3}{5}</math>. It is known that <math>\sin(x)=\cos(90-x)</math>, so <math>180-C=90-(180-C)=180-B</math>. We simplify this as follows:
+<cmath>-90+C=180-B</cmath>
+<cmath>B+C=270^{\circ}</cmath>
+Since <math>B+C=270^{\circ}</math>, we know that <math>A+D=360-(B+C)=90^{\circ}</math>. Now extend <math>AB</math> and <math>CD</math> 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 <math>AB</math> and <math>CD</math> intersect at <math>E</math>. We know that <math>\angle AED=90^{\circ}</math> because <math>\angle E = 180 - (A+D)=180-90 = 90^{\circ}</math>.
+Since <math>\sin BCD = \frac{3}{5}</math>, we get <math>\sin ECB=\sin(180-BCD)=\sin BCD = \frac{3}{5}</math>. Thus, <math>EB=3</math> and <math>EC=4</math> from simple sin application.
+<math>AD</math> is the hypotenuse of right <math>\triangle AED</math>, with leg lengths <math>AB+BE=7</math> and <math>EC+CD=24</math>. Thus, <math>AD=\boxed{\textbf{(E)}25}</math>
+-WannabeCharmander
 == See also ==