Difference between revisions of "1999 AMC 8 Problems/Problem 21"
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==See Also== | ==See Also== | ||
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Revision as of 23:34, 4 July 2013
Problem
The degree measure of angle 
is
![[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); [/asy]](http://latex.artofproblemsolving.com/c/c/3/cc3d87ddc15f303c48dcb5e245523d226f95fbe3.png)
Solution
Angle-chasing using the small triangles:
Use the line below and to the left of the 
angle to find that the rightmost angle in the small lower-left triangle is 
.
Then use the small lower-left triangle to find that the remaining angle in that triangle is 
.
Use congruent vertical angles to find that the lower angle in the smallest triangle containing is also 
.
Next, use line segment 
to find that the other angle in the smallest triangle contianing is 
.
The small triangle containing has a angle and an 
angle. The remaining angle must be 
See Also
| 1999 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
| 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 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
