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

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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>
 
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>
  
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== Solution 2 ==
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== Solution 2 ==
 
  
 
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} 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 {APE} is an isosceles triangle, we get that </math>\angle{EAP} = (180 - 110)/2 = 35<math>. Therefore, </math>\angle{EAP} = 35$.
 
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} 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 {APE} is an isosceles triangle, we get that </math>\angle{EAP} = (180 - 110)/2 = 35<math>. Therefore, </math>\angle{EAP} = 35$.

Revision as of 15:48, 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 $\boxed{35 \text{(B)}}$

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,$\angle{EAP} = 35$.

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