Difference between revisions of "2014 AMC 10A Problems/Problem 22"

Revision as of 15:36, 18 February 2014

Problem

In rectangle $ABCD$ , $AB=20$ and $BC=10$ . Let $E$ be a point on $\overline{CD}$ such that $\angle CBE=15^\circ$ . What is $AE$ ?

$\textbf{(A)}\ \dfrac{20\sqrt3}3\qquad\textbf{(B)}\ 10\sqrt3\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 11\sqrt3\qquad\textbf{(E)}\ 20$

Solution

Note that $\tan 15^\circ=\frac{EC}{10} \Rightarrow EC=20-10 \sqrt 3$ . Therefore, we have $DE=10\sqrt 3$ . Since $ADE$ is a $30-60-90$ triangle, $AE=2 \cdot DE=2 \cdot 10=\boxed{\textbf{(E)} \: 20}$

2014 AMC 10A (Problems • Answer Key • Resources)
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 ==Solution==
-Note that <math>\tan 15^\circ=\frac{EC}{10} \Rightarrow EC=20-10 \sqrt 3</math>. Therefore, we have <math>DC=10\sqrt 3</math>. Since <math>ADE</math> is a <math>30-60-90</math> triangle, <math>AE=2 \cdot DE=2 \cdot 10=\boxed{\textbf{(E)} \: 20}</math>
+Note that <math>\tan 15^\circ=\frac{EC}{10} \Rightarrow EC=20-10 \sqrt 3</math>. Therefore, we have <math>DE=10\sqrt 3</math>. Since <math>ADE</math> is a <math>30-60-90</math> triangle, <math>AE=2 \cdot DE=2 \cdot 10=\boxed{\textbf{(E)} \: 20}</math>
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Difference between revisions of "2014 AMC 10A Problems/Problem 22"

Revision as of 15:36, 18 February 2014

Problem

Solution

See Also