2016 AMC 12B Problem 13

Question

Alice and Bob live

10

miles apart. One day Alice looks due north from her house and sees an airplane. At the same time Bob looks due west from his house and sees the same airplane. The angle of elevation of the airplane is

30^\circ

from Alice’s position and

60^\circ

from Bob’s position. Which of the following is closest to the airplane’s altitude, in miles?

3.5

4

4.5

5

5.5

Answer: E

Concepts:special right triangle Pythagorean Theorem 3D geometry

Difficulty rating: 1630

Small hint:

Let the plane be above point

D

on the ground at height

h.

Triangles

ACD

and

BCD

are

30

-

60

-

90

right triangles

Big hint:

AD=\sqrt3\,h

and

BD=\dfrac{h}{\sqrt3};

then

AD^2+BD^2=10^2

Solution:

Let the airplane be at

C,

directly above point

D

on the ground at altitude

h.

Triangles

ACD

and

BCD

are

30

-

60

-

90

right triangles, so

AD=\sqrt3\,h

and

BD=\dfrac{h}{\sqrt3}.

Since Alice looks north and Bob looks west,

\angle ADB=90^\circ,

so

AD^2+BD^2=AB^2=100.

Then

3h^2+\dfrac{h^2}{3}=\dfrac{10h^2}{3}=100,

giving

h=\sqrt{30}\approx5.48,

closest to

5.5.

Thus, the correct answer is E.