Difference between revisions of "1999 AMC 8 Problems/Problem 21"
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==Solution== | ==Solution== | ||
+ | ===Solution 1=== | ||
Angle-chasing using the small triangles: | Angle-chasing using the small triangles: | ||
Line 27: | Line 28: | ||
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>. | ||
− | Next, use line segment <math>AB</math> to find that the other angle in the smallest triangle | + | 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>. | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/suaYxFnoU6E?t=99 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | == Video Solution by CosineMethod [🔥Fast and Easy🔥]== | ||
+ | |||
+ | https://youtu.be/X_SbFalrsV8?si=BLdIghVBvHNMmEIn | ||
==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}} |
Revision as of 22:27, 24 January 2024
Contents
Problem
The degree measure of angle is
Solution
Solution 1
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 containing
is
.
The small triangle containing has a
angle and an
angle. The remaining angle must be
Solution 2
The third angle of the triangle containing the angle and the
angle is
. It follows that
is the third angle of the triangle consisting of the found
angle and the given
angle. Thus,
is a
angle, and so the answer is
.
Video Solution by OmegaLearn
https://youtu.be/suaYxFnoU6E?t=99
~ pi_is_3.14
Video Solution by CosineMethod [🔥Fast and Easy🔥]
https://youtu.be/X_SbFalrsV8?si=BLdIghVBvHNMmEIn
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.