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Difference between revisions of "1999 AMC 8 Problems/Problem 21"
(Solution 2)
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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==
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==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 $A$ 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]

$\text{(A)}\ 20 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 35 \qquad \text{(D)}\ 40 \qquad \text{(E)}\ 45$

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] Note that $\angle B=180-100-40=40^\circ$. So $\angle A=180-110-40=\boxed{30^\circ}$.

Solution 2

Angle-chasing using the small triangles:

Use the line below and to the left of the $110^\circ$ angle to find that the rightmost angle in the small lower-left triangle is $180 - 110 = 70^\circ$.

Then use the small lower-left triangle to find that the remaining angle in that triangle is $180 - 70 - 40 = 70^\circ$.

Use congruent vertical angles to find that the lower angle in the smallest triangle containing $A$ is also $70^\circ$.

Next, use line segment $AB$ to find that the other angle in the smallest triangle contianing $A$ is $180 - 100 = 80^\circ$.

The small triangle containing $A$ has a $70^\circ$ angle and an $80^\circ$ angle. The remaining angle must be $180 - 70 - 80 = \boxed{30^\circ, B}$

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