Difference between revisions of "2009 AMC 8 Problems/Problem 19"
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Adding up all the cases, we get <math>70+55+40=165</math>, so the answer is <math> \boxed{\textbf{(D)}\ 165}</math>. | Adding up all the cases, we get <math>70+55+40=165</math>, so the answer is <math> \boxed{\textbf{(D)}\ 165}</math>. | ||
− | == Video Solution == | + | == Video Solution by OmegaLearn == |
https://youtu.be/FDgcLW4frg8?t=902 | https://youtu.be/FDgcLW4frg8?t=902 | ||
~ pi_is_3.14 | ~ pi_is_3.14 | ||
+ | https://www.youtube.com/watch?v=EO3DphZAlDY ~David | ||
==Video Solution 2== | ==Video Solution 2== | ||
https://youtu.be/iB0dcpVREE8 Soo, DRMS, NM | https://youtu.be/iB0dcpVREE8 Soo, DRMS, NM | ||
+ | |||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2009|num-b=18|num-a=20}} | {{AMC8 box|year=2009|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 18:38, 15 April 2023
Problem
Two angles of an isosceles triangle measure and . What is the sum of the three possible values of ?
Solution
There are 3 cases: where is a base angle with the as the other angle, where is a base angle with as the vertex angle, and where is the vertex angle with as a base angle.
Case 1: is a base angle with the as the other angle: Here, , since base angles are congruent.
Case 2: is a base angle with as the vertex angle: Here, the 2 base angles are both , so we can use the equation , which simplifies to .
Case 3: is the vertex angle with as a base angle: Here, both base angles are , since base angles are congruent. Thus, we can use the equation , which simplifies to .
Adding up all the cases, we get , so the answer is .
Video Solution by OmegaLearn
https://youtu.be/FDgcLW4frg8?t=902
~ pi_is_3.14
https://www.youtube.com/watch?v=EO3DphZAlDY ~David
Video Solution 2
https://youtu.be/iB0dcpVREE8 Soo, DRMS, NM
See Also
2009 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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.