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

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

Revision as of 16:31, 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). 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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