Difference between revisions of "2000 AMC 8 Problems/Problem 24"
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+ | ==Problem== | ||
+ | |||
If <math> \angle A = 20^\circ </math> and <math> \angle AFG =\angle AGF </math>, then <math> \angle B+\angle D = </math> | If <math> \angle A = 20^\circ </math> and <math> \angle AFG =\angle AGF </math>, then <math> \angle B+\angle D = </math> | ||
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<math> \text{(A)}\ 48^\circ\qquad\text{(B)}\ 60^\circ\qquad\text{(C)}\ 72^\circ\qquad\text{(D)}\ 80^\circ\qquad\text{(E)}\ 90^\circ </math> | <math> \text{(A)}\ 48^\circ\qquad\text{(B)}\ 60^\circ\qquad\text{(C)}\ 72^\circ\qquad\text{(D)}\ 80^\circ\qquad\text{(E)}\ 90^\circ </math> | ||
+ | |||
+ | ==Solution== | ||
+ | |||
+ | As a strategy, think of how <math>\angle B + \angle D</math> would be determined, particularly without determining either of the angles individually, since it may not be possible to determine <math>\angle B</math> or <math>\angle D</math> alone. If you see <math>\triangle BFD</math>, the you can see that the problem is solved quickly after determining <math>\angle BFD</math>. | ||
+ | |||
+ | But start with <math>\triangle AGF</math>, since that's where most of our information is. Looking at <math>\triangle AGF</math>, since <math>\angle F = \angle G</math>, and <math>\angle A = 20</math>, we can write: | ||
+ | |||
+ | <math>\angle A + \angle G + \angle F = 180</math> | ||
+ | |||
+ | <math>20 + 2\angle F = 180</math> | ||
+ | |||
+ | <math>\angle AFG = 80</math> | ||
+ | |||
+ | 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>80 + \angle GFD = 180</math> | ||
+ | |||
+ | <math>\angle GFD = 100</math> | ||
+ | |||
+ | Ignoring all other parts of the figure and looking only at <math>\triangle BFD</math>, you see that <math>\angle B + \angle D + \angle F = 180</math>. But <math>\angle F</math> is the same as <math>\angle GFD</math>. Therefore: | ||
+ | |||
+ | <math>\angle B + \angle D + \angle GFD = 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> | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{AMC8 box|year=2000|num-b=23|num-a=25}} | ||
+ | {{MAA Notice}} |
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 |
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