2014 AMC 10A Problems/Problem 22

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, $AE=2 \cdot AD=2 \cdot 10=\boxed{\textbf{(E)} \: 20}$

2014 AMC 10A (Problems • Answer Key • Resources)
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2014 AMC 10A Problems/Problem 22

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