Difference between revisions of "2014 AMC 8 Problems/Problem 9"
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<math>\textbf{(A) }100\qquad\textbf{(B) }120\qquad\textbf{(C) }135\qquad\textbf{(D) }140\qquad \textbf{(E) }150</math> | <math>\textbf{(A) }100\qquad\textbf{(B) }120\qquad\textbf{(C) }135\qquad\textbf{(D) }140\qquad \textbf{(E) }150</math> | ||
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+ | ==Video Solution (CREATIVE THINKING)== | ||
+ | https://youtu.be/jLnqUOe0HPE | ||
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+ | ~Education, the Study of Everything | ||
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==Video Solution== | ==Video Solution== | ||
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https://youtu.be/j5KrHM81HZ8 ~savannahsolver | https://youtu.be/j5KrHM81HZ8 ~savannahsolver | ||
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https://youtu.be/abSgjn4Qs34?t=3140 | https://youtu.be/abSgjn4Qs34?t=3140 | ||
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==Solution== | ==Solution== |
Latest revision as of 11:14, 2 July 2023
Contents
Problem
In , is a point on side such that and measures . What is the degree measure of ?
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution
https://www.youtube.com/watch?v=HP-lBKohxhE ~David
https://youtu.be/j5KrHM81HZ8 ~savannahsolver
Video Solution
https://youtu.be/abSgjn4Qs34?t=3140
Solution
Using angle chasing is a good way to solve this problem. , so . Then . Since and are supplementary, .
See Also
2014 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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.