1999 AMC 8 Problems/Problem 21: Difference between revisions
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<math>\text{(A)}\ 20 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 35 \qquad \text{(D)}\ 40 \qquad \text{(E)}\ 45</math> | <math>\text{(A)}\ 20 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 35 \qquad \text{(D)}\ 40 \qquad \text{(E)}\ 45</math> | ||
==Solution== | ==Solution 1== | ||
<asy> | <asy> | ||
unitsize(12); | unitsize(12); | ||
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</asy> | </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>. | ||
==Solution 2== | |||
Angle-chasing using the small triangles: | |||
Use the line below and to the left of the <math>110^\circ</math> angle to find that the rightmost angle in the small lower-left triangle is <math>180 - 110 = 70^\circ</math>. | |||
Then use the small lower-left triangle to find that the remaining angle in that triangle is <math>180 - 70 - 40 = 70^\circ</math>. | |||
Use congruent vertical angles to find that the lower angle in the smallest triangle containing <math>A</math> is also <math>70^\circ</math>. | |||
Next, use line segment <math>AB</math> to find that the other angle in the smallest triangle contianing <math>A</math> is <math>180 - 100 = 80^\circ</math>. | |||
The small triangle containing <math>A</math> has a <math>70^\circ</math> angle and an <math>80^\circ</math> angle. The remaining angle must be <math>180 - 70 - 80 = \boxed{30^\circ, B}</math> | |||
Revision as of 16:14, 30 July 2011
Problem 21
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 1
![[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), NE); [/asy]](http://latex.artofproblemsolving.com/f/3/5/f35551ea4d828905a2f3e1d6a2abcc033d9b8d36.png)
Note that 
. So 
.
Solution 2
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 
