Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

1979 AHSME Problems/Problem 24: Difference between revisions

← Older edit
Wannabecharmander (talk | contribs)
editor
38 edits
Wannabecharmander (talk | contribs)
editor
38 edits
 
(2 intermediate revisions by the same user not shown)
Line 38: Line 38:
draw((0,4)--(0,7), dashed);
draw((0,4)--(0,7), dashed);
draw((0,7)--(4,7), dashed);
draw((0,7)--(4,7), dashed);
//diagram by WannabeCharmander
</asy>
</asy>


Let <math>AB</math> and <math>CD</math> intersect at <math>E</math>.
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>.


<math>\boxed{\textbf{E}}</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 ==
== See also ==

Latest revision as of 22:23, 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

See also

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1979_AHSME_Problems/Problem_24&oldid=95493"