Question

Rectangle $ABCD$ has $AB = 6$ and $BC = 3.$ Point $M$ is chosen on side $AB$ so that $\angle AMD = \angle CMD.$ What is the degree measure of $\angle AMD?$

$15$

$30$

$45$

$60$

$75$

Answer: E

Concepts:rectangle isosceles triangle trigonometry

Difficulty rating: 1540

Small Hint:

Use $AB\parallel CD$ to transfer one angle to $\angle CDM$

Big Hint:

Triangle $CMD$ becomes isosceles

Solution:

The angles $\angle AMD$ and $\angle MDC$ are equal since $AB \parallel DC.$

As such, $\angle MDC = \angle DMC ,$ making $MDC$ isosceles and $MC = DC = 6.$

As we can see, $\sin (\angle CMB) = \frac 12,$ making $\angle CMB = 30^\circ .$

Therefore, $\angle AMC = 150^\circ .$ Since $\angle AMD$ is half of that, $\angle AMD = 75^\circ .$

Thus, the correct answer is E .

2011 AMC 10B Problem 18

Answer: E

Difficulty rating: 1540

Small Hint:

Big Hint:

Solution:

Problem 18 in Other Years