Difference between revisions of "2000 AMC 8 Problems/Problem 24"
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<math>\angle AFG = 80</math> | <math>\angle AFG = 80</math> | ||
− | By noting that <math>\angle AFG</math> and <math>\angle GFD</math> make a | + | By noting that <math>\angle AFG</math> and <math>\angle GFD</math> make a straight line, we know |
<math>\angle AFG + \angle GFD = 180</math> | <math>\angle AFG + \angle GFD = 180</math> | ||
Line 48: | Line 48: | ||
<math>\angle B + \angle D + 100 = 180</math> | <math>\angle B + \angle D + 100 = 180</math> | ||
<math>\angle B + \angle D = 80^\circ</math>, and the answer is thus <math>\boxed{D}</math> | <math>\angle B + \angle D = 80^\circ</math>, and the answer is thus <math>\boxed{D}</math> | ||
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==See Also== | ==See Also== |
Latest revision as of 07:25, 14 October 2015
Problem
If and , then
Solution
As a strategy, think of how would be determined, particularly without determining either of the angles individually, since it may not be possible to determine or alone. If you see , the you can see that the problem is solved quickly after determining .
But start with , since that's where most of our information is. Looking at , since , and , we can write:
By noting that and make a straight line, we know
Ignoring all other parts of the figure and looking only at , you see that . But is the same as . Therefore:
, and the answer is thus
See Also
2000 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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.