Difference between revisions of "2021 Fall AMC 10A Problems/Problem 7"
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~MRENTHUSIASM | ~MRENTHUSIASM | ||
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+ | ==Solution 2 (same as Solution 1 but by another user)== | ||
+ | Since <math>\angle CDE=110^\circ</math> and <math>\angle ADC=90^\circ,</math> we know that <math>\angle FDE=360^\circ-110^\circ-90^\circ=160^\circ.</math> From the given, <math>DE=DF,</math> so that means <math>\triangle DEF</math> is isosceles. So, <math>\angle EFD=\left(\frac{180-160}{2}\right)^\circ=10^\circ.</math> Hence, <math>\angle AFE=180^\circ-10^\circ=170^\circ</math> since it is supplementary to <math>\angle EFD.</math> This gives us answer choice <math>\boxed{\textbf{(D)}}.</math> | ||
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+ | [[User:Aops-g5-gethsemanea2|Aops-g5-gethsemanea2]] ([[User talk:Aops-g5-gethsemanea2|talk]]) 18:56, 22 November 2021 (EST) | ||
+ | This solution conflicted with the above solution due to two people editing at the same time. | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2021 Fall|ab=A|num-b=6|num-a=8}} | {{AMC10 box|year=2021 Fall|ab=A|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:56, 22 November 2021
Contents
[hide]Problem
As shown in the figure below, point lies on the opposite half-plane determined by line from point so that . Point lies on so that , and is a square. What is the degree measure of ?
Solution
By angle subtraction, we have
Note that is isosceles, so Finally, we get degrees.
~MRENTHUSIASM
Solution 2 (same as Solution 1 but by another user)
Since and we know that From the given, so that means is isosceles. So, Hence, since it is supplementary to This gives us answer choice
Aops-g5-gethsemanea2 (talk) 18:56, 22 November 2021 (EST) This solution conflicted with the above solution due to two people editing at the same time.
See Also
2021 Fall AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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 AMC 10 Problems and Solutions |
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