2009 AMC 12A Problem 13

Question

A ship sails

10

miles in a straight line from

A

to

B,

turns through an angle between

45^\circ

and

60^\circ,

and then sails another

20

miles to

C.

Let

AC

be measured in miles. Which of the following intervals contains

AC^2?

[400, 500]

[500, 600]

[600, 700]

[700, 800]

[800, 900]

By the Law of Cosines,

AC^2 = 10^2

+ 20^2

- 2\cdot 10\cdot 20\cos(\angle ABC)

The turn is between

45^\circ

and

60^\circ,

so

\angle ABC

is between

120^\circ

and

135^\circ

By the Law of Cosines,

\begin{aligned} AC^2 &= 10^2 + 20^2 \\ &\quad {}- 2\cdot 10\cdot 20\cos(\angle ABC) \\ &= 500 - 400\cos(\angle ABC). \end{aligned}

The ship turns through an angle between

45^\circ

and

60^\circ,

so the interior angle

\angle ABC

lies between

120^\circ

and

135^\circ.

Since

\cos 120^\circ = -\dfrac{1}{2}

and

\cos 135^\circ = -\dfrac{\sqrt{2}}{2},

\begin{aligned} 700 &= 500 + 200 \\ &\le AC^2 \le 500 + 200\sqrt{2} \\ &\lt 800. \end{aligned}

So

AC^2

lies in

[700, 800].

Thus, the correct answer is D.

Problem 13 in other years