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

1999 AMC 8 Problems/Problem 21: Difference between revisions

Newer edit →
Hydroxide (talk | contribs)
31 edits
Created page with "[asy] unitsize(12); draw((0,0)--(20,0)--(1,-10)--(9,5)--(18,-8)--cycle); draw(arc((1,-10),(1+19/sqrt(461),-10+10/sqrt(461)),(25/17,-155/17),CCW)); draw(arc((19/3,0),(19/3-8/17,-1..."
 
Hydroxide (talk | contribs)
31 edits
No edit summary
Line 1: Line 1:
[asy]
<asy>
unitsize(12);
unitsize(12);
draw((0,0)--(20,0)--(1,-10)--(9,5)--(18,-8)--cycle);
draw((0,0)--(20,0)--(1,-10)--(9,5)--(18,-8)--cycle);
Line 5: Line 5:
draw(arc((19/3,0),(19/3-8/17,-15/17),(22/3,0),CCW));
draw(arc((19/3,0),(19/3-8/17,-15/17),(22/3,0),CCW));
draw(arc((900/83,-400/83),(900/83+19/sqrt(461),-400/83+10/sqrt(461)),(900/83 - 9/sqrt(97),-400/83 + 4/sqrt(97)),CCW));
draw(arc((900/83,-400/83),(900/83+19/sqrt(461),-400/83+10/sqrt(461)),(900/83 - 9/sqrt(97),-400/83 + 4/sqrt(97)),CCW));
label(rotate(30)*"<math>40^\circ</math>",(2,-8.9),ENE);
label(rotate(30)*"$40^\circ$",(2,-8.9),ENE);
label("<math>100^\circ</math>",(21/3,-2/3),SE);
label("$100^\circ$",(21/3,-2/3),SE);
label("<math>110^\circ</math>",(900/83,-317/83),NNW);
label("$110^\circ$",(900/83,-317/83),NNW);
label("<math>A</math>",(0,0),NW);
label("$A$",(0,0),NW);
label("<math>B</math>", (20,0), NW);[/asy]
label("$B$", (20,0), NW);
</asy>
Note that <math>\angle B=180-100-40=40^\circ</math>. So <math>\angle A=180-110-40=\boxed{30^\circ}</math>.
Note that <math>\angle B=180-100-40=40^\circ</math>. So <math>\angle A=180-110-40=\boxed{30^\circ}</math>.

Revision as of 11:57, 17 June 2011

[asy] unitsize(12); draw((0,0)--(20,0)--(1,-10)--(9,5)--(18,-8)--cycle); draw(arc((1,-10),(1+19/sqrt(461),-10+10/sqrt(461)),(25/17,-155/17),CCW)); draw(arc((19/3,0),(19/3-8/17,-15/17),(22/3,0),CCW)); draw(arc((900/83,-400/83),(900/83+19/sqrt(461),-400/83+10/sqrt(461)),(900/83 - 9/sqrt(97),-400/83 + 4/sqrt(97)),CCW)); label(rotate(30)*"$40^\circ$",(2,-8.9),ENE); label("$100^\circ$",(21/3,-2/3),SE); label("$110^\circ$",(900/83,-317/83),NNW); label("$A$",(0,0),NW); label("$B$", (20,0), NW); [/asy] Note that $\angle B=180-100-40=40^\circ$. So $\angle A=180-110-40=\boxed{30^\circ}$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1999_AMC_8_Problems/Problem_21&oldid=39651"