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Difference between revisions of "2023 AMC 10B Problems/Problem 7"
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== Solution 1==
 
== Solution 1==
  
−
First, let's call the center of both squares <math>I</math>. Then, <math>\angle{AIE} = 20</math>, and since <math>\overline{EI} = \overline{AI}</math>, <math>\angle{AEI} = \angle{EAI} = 80</math>. Then, we know that <math>AI</math> bisects angle <math>\angle{DAB}</math>, so <math>\angle{BAI} = \angle{DAI} = 45</math>. Subtracting <math>45</math> from <math>80</math>, we get <math>\fbox{\bf{35}}</math>
+
First, let's call the center of both squares <math>I</math>. Then, <math>\angle{AIE} = 20</math>, and since <math>\overline{EI} = \overline{AI}</math>, <math>\angle{AEI} = \angle{EAI} = 80</math>. Then, we know that <math>AI</math> bisects angle <math>\angle{DAB}</math>, so <math>\angle{BAI} = \angle{DAI} = 45</math>. Subtracting <math>45</math> from <math>80</math>, we get <math>\boxed{35 \text{B}}</math>

Revision as of 16:32, 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 $\boxed{35 \text{B}}$

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