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

Revision as of 13:01, 26 December 2015

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 $\tan 15^\circ=\frac{EC}{10} \Rightarrow EC=20-10 \sqrt 3$ . (If you do not know the tangent half-angle formula, it is $\frac{1-\cos a}{\sin a}$ ). Therefore, we have $DE=10\sqrt 3$ . Since $ADE$ is a $30-60-90$ triangle, $AE=2 \cdot AD=2 \cdot 10=\boxed{\textbf{(E)} \: 20}$

Solution 2

Mark point $F$ on line $CD$ such that

2014 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 10: / Line 10: @@
 ==Solution 2==
+Mark point <math>F</math> on line <math>CD</math> such that
 ==See Also==

Page

Toolbox

Search

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

Revision as of 13:01, 26 December 2015

Contents

Problem

Solution

Solution 2

See Also