Difference between revisions of "2009 AMC 8 Problems/Problem 19"
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Pi is 3.14 (talk | contribs) (→Video Solution) |
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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 | ||
Revision as of 04:25, 16 January 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
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.