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Difference between revisions of "2023 AMC 10B Problems/Problem 7"

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Sqrt <math>ABCD</math> is rotated <math>20^{\circ}</math> clockwise about its center to obtain square <math>EFGH</math>, as shown below(Please help me add diagram and then remove this). What is the degree measure of <math>\angle EAB</math>?
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Sqrt <math>ABCD</math> is rotated <math>20^{\circ}</math> clockwise about its center to obtain square <math>EFGH</math>, as shown below(Please help me add diagram and then remove this).  
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<img src="https://wiki-images.artofproblemsolving.com//9/9d/IMG_1031.jpeg" alt="IMG 1031.jpeg"/>
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What is the degree measure of <math>\angle EAB</math>?
  
 
<math>\text{(A)}\ 24^{\circ} \qquad \text{(B)}\ 35^{\circ} \qquad \text{(C)}\ 30^{\circ} \qquad \text{(D)}\ 32^{\circ} \qquad \text{(E)}\ 20^{\circ}</math>
 
<math>\text{(A)}\ 24^{\circ} \qquad \text{(B)}\ 35^{\circ} \qquad \text{(C)}\ 30^{\circ} \qquad \text{(D)}\ 32^{\circ} \qquad \text{(E)}\ 20^{\circ}</math>

Revision as of 19:13, 15 November 2023

Sqrt $ABCD$ is rotated $20^{\circ}$ clockwise about its center to obtain square $EFGH$, as shown below(Please help me add diagram and then remove this). <img src="https://wiki-images.artofproblemsolving.com//9/9d/IMG_1031.jpeg" alt="IMG 1031.jpeg"/> What is the degree measure of $\angle EAB$?

$\text{(A)}\ 24^{\circ} \qquad \text{(B)}\ 35^{\circ} \qquad \text{(C)}\ 30^{\circ} \qquad \text{(D)}\ 32^{\circ} \qquad \text{(E)}\ 20^{\circ}$

Solution 1

First, let's call the center of both squares $I$. Then, $\angle{AIE} = 20$, and since $\overline{EI} = \overline{AI}$, $\angle{AEI} = \angle{EAI} = 80$. Then, we know that $AI$ bisects angle $\angle{DAB}$, so $\angle{BAI} = \angle{DAI} = 45$. Subtracting $45$ from $80$, we get $\boxed{\text{(B)} 35}$

~jonathanzhou18

Solution 2

First, label the point between $A$ and $H$ point $O$ and the point between $A$ and $H$ point $P$. We know that $\angle{AOP} = 20$ and that $\angle{A} = 90$. Subtracting $20$ and $90$ from $180$, we get that $\angle{APO}$ is $70$. Subtracting $70$ from $180$, we get that $\angle{OPB} = 110$. From this, we derive that $\angle{APE} = 110$. Since triangle $APE$ is an isosceles triangle, we get that $\angle{EAP} = (180 - 110)/2 = 35$. Therefore, $\angle{EAB} = 35$. The answer is $\boxed{\text{(B)} 35}$.

~yourmomisalosinggame (a.k.a. Aaron)

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