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Difference between revisions of "1979 AHSME Problems/Problem 3"

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Notice that <math>\measuredangle DAE = 90^\circ+60^\circ = 150^\circ</math> and that <math>AD = AE</math>. Then triangle <math>ADE</math> is isosceles, so <math>\measuredangle AED = \dfrac{180^\circ-150^\circ}{2} = \boxed{\textbf{(C) } 15}</math>.
 
Notice that <math>\measuredangle DAE = 90^\circ+60^\circ = 150^\circ</math> and that <math>AD = AE</math>. Then triangle <math>ADE</math> is isosceles, so <math>\measuredangle AED = \dfrac{180^\circ-150^\circ}{2} = \boxed{\textbf{(C) } 15}</math>.
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== Solution 2 (Obnoxiously tedious) ==
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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>.
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Please don't try this solution. This is just obnoxiously tedious and impractical.
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~hastapasta
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== Video Solution by OmegaLearn==
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https://youtu.be/FDgcLW4frg8?t=1341
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~ pi_is_3.14
  
 
== See also ==
 
== See also ==
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{{AHSME box|year=1979|before=num-b=2|num-a=4}}
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{{AHSME box|year=1979|num-b=2|num-a=4}}
  
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[[Category:Introductory Algebra Problems]]
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[[Category:Introductory Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 23:42, 31 December 2022

Problem 3

[asy] real s=sqrt(3)/2; draw(box((0,0),(1,1))); draw((1+s,0.5)--(1,1)); draw((1+s,0.5)--(1,0)); draw((0,1)--(1+s,0.5)); label("$A$",(1,1),N); label("$B$",(1,0),S); label("$C$",(0,0),W); label("$D$",(0,1),W); label("$E$",(1+s,0.5),E); //Credit to TheMaskedMagician for the diagram[/asy]

In the adjoining figure, $ABCD$ is a square, $ABE$ is an equilateral triangle and point $E$ is outside square $ABCD$. What is the measure of $\measuredangle AED$ in degrees?

$\textbf{(A) }10\qquad \textbf{(B) }12.5\qquad \textbf{(C) }15\qquad \textbf{(D) }20\qquad \textbf{(E) }25$

Solution

Solution by e_power_pi_times_i

Notice that $\measuredangle DAE = 90^\circ+60^\circ = 150^\circ$ and that $AD = AE$. Then triangle $ADE$ is isosceles, so $\measuredangle AED = \dfrac{180^\circ-150^\circ}{2} = \boxed{\textbf{(C) } 15}$.

Solution 2 (Obnoxiously tedious)

WLOG, let the side length of the square and the equilateral triangle be $1$. $\angle{DAE}=90^\circ+60^\circ=150^\circ$. Apply the law of cosines then the law of sines, we find that $\angle{AED}=15^\circ$. Select $\boxed{C}$.

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

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