Difference between revisions of "2021 Fall AMC 10A Problems/Problem 7"
MRENTHUSIASM (talk | contribs) (→Solution 2 (same as Solution 1 but by another user): To clean up spaces, I will delete Sol 2 and give credit to both users.) |
MRENTHUSIASM (talk | contribs) m (→Solution) |
||
Line 32: | Line 32: | ||
Note that <math>\triangle DEF</math> is isosceles, so <math>\angle EFD = \frac{180^\circ - \angle ADE}{2}=10^\circ.</math> Finally, we get <math>\angle AFE = 180^\circ - \angle EFD = \boxed{\textbf{(D) }170}</math> degrees. | Note that <math>\triangle DEF</math> is isosceles, so <math>\angle EFD = \frac{180^\circ - \angle ADE}{2}=10^\circ.</math> Finally, we get <math>\angle AFE = 180^\circ - \angle EFD = \boxed{\textbf{(D) }170}</math> degrees. | ||
− | ~MRENTHUSIASM | + | ~MRENTHUSIASM ~[[User:Aops-g5-gethsemanea2|Aops-g5-gethsemanea2]] |
==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 18:59, 22 November 2021
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 ~Aops-g5-gethsemanea2
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.