Question

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

$\dfrac{20\sqrt3}3$

$10\sqrt3$

$18$

$11\sqrt3$

$20$

Answer: E

Concepts:special right triangle isosceles triangle angle chasing

Difficulty rating: 1950

Solution:

Let $E'$ be the point on $\overline{CD}$ such that $AE'=AB=20$ .

Since $AD=10$ , triangle $ADE'$ is a $30$ - $60$ - $90$ triangle, so $\angle DAE'=60^\circ$ and $\angle BAE'=30^\circ$ .

Also $AE'=AB$ , so triangle $ABE'$ is isosceles. Its vertex angle at $A$ is $30^\circ$ , so each base angle is $75^\circ$ .

Therefore $\angle CBE'=90^\circ-75^\circ=15^\circ$ , so $E'=E$ , and $AE=20$ .

Thus, E is the correct answer.

2014 AMC 10A Problem 22

Answer: E

Difficulty rating: 1950

Solution:

Problem 22 in Other Years