Difference between revisions of "1979 AHSME Problems/Problem 3"
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WLOG, let the side length of the square and the equilateral triangle be <math>1</math>. <math>\angle{DAE}=90^\circ+60^\circ=150^\circ</math>. Apply the law of cosines then the law of sines, we find that <math>\angle{AED}=15^\circ</math>. Select <math>\boxed{C}</math>. | WLOG, let the side length of the square and the equilateral triangle be <math>1</math>. <math>\angle{DAE}=90^\circ+60^\circ=150^\circ</math>. Apply the law of cosines then the law of sines, we find that <math>\angle{AED}=15^\circ</math>. Select <math>\boxed{C}</math>. | ||
+ | |||
+ | Please don't try this solution. This is just obnoxiously tedious and impractical. | ||
~hastapasta | ~hastapasta | ||
− | == Video Solution == | + | |
+ | == Video Solution by OmegaLearn== | ||
https://youtu.be/FDgcLW4frg8?t=1341 | https://youtu.be/FDgcLW4frg8?t=1341 | ||
Latest revision as of 00:42, 1 January 2023
Contents
Problem 3
In the adjoining figure, is a square,
is an equilateral triangle and point
is outside square
.
What is the measure of
in degrees?
Solution
Solution by e_power_pi_times_i
Notice that and that
. Then triangle
is isosceles, so
.
Solution 2 (Obnoxiously tedious)
WLOG, let the side length of the square and the equilateral triangle be .
. Apply the law of cosines then the law of sines, we find that
. Select
.
Please don't try this solution. This is just obnoxiously tedious and impractical.
~hastapasta
Video Solution by OmegaLearn
https://youtu.be/FDgcLW4frg8?t=1341
~ pi_is_3.14
See also
1979 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.