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Difference between revisions of "1999 AMC 8 Problems/Problem 21"
(→‎Solution)
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==Solution==
 
==Solution==
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===Solution 1===
  
 
Angle-chasing using the small triangles:
 
Angle-chasing using the small triangles:
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Use congruent vertical angles to find that the lower angle in the smallest triangle containing <math>A</math> is also <math>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>.   
  
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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>.
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Next, use line segment <math>AB</math> to find that the other angle in the smallest triangle containing <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>
 
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>
 +
 +
===Solution 2===
 +
The third angle of the triangle containing the <math>100^\circ</math> angle and the <math>40^\circ</math> angle is <math>180^\circ - 100^\circ - 40^\circ = 40^\circ</math>. It follows that <math>A</math> is the third angle of the triangle consisting of the found <math>40^\circ</math> angle and the given <math>110^\circ</math> angle. Thus, <math>A</math> is a <math>180^\circ - 110^\circ - 40^\circ = 30^\circ</math> angle, and so the answer is <math>\boxed{30^\circ, \textbf{B}}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=1999|num-b=20|num-a=22}}
 
{{AMC8 box|year=1999|num-b=20|num-a=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 11:56, 27 September 2015

Problem

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

Solution 1

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 containing $A$ 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 $180 - 70 - 80 = \boxed{30^\circ, B}$

Solution 2

The third angle of the triangle containing the $100^\circ$ angle and the $40^\circ$ angle is $180^\circ - 100^\circ - 40^\circ = 40^\circ$. It follows that $A$ is the third angle of the triangle consisting of the found $40^\circ$ angle and the given $110^\circ$ angle. Thus, $A$ is a $180^\circ - 110^\circ - 40^\circ = 30^\circ$ angle, and so the answer is $\boxed{30^\circ, \textbf{B}}$.

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

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