A security camera is mounted on a vertical wall at point C, exactly 15 feet above the flat… — ACT

A security camera is mounted on a vertical wall at point $C$ , exactly $15$ feet above the flat ground. The camera is programmed to monitor two objects, $A$ and $B$ , on the ground. The line of sight from the camera to object $A$ makes a $30^\circ$ angle with the wall, and the line of sight from the camera to object $B$ makes a $45^\circ$ angle with the wall. On the ground, the path from the base of the wall directly below the camera to object $A$ is perpendicular to the path from the base of the wall to object $B$ . What is the straight-line distance, in feet, between object $A$ and object $B$ ?

A
$5\sqrt{6}$
B
$4\sqrt{15}$
$10\sqrt{3}$ Answer
D
$15 + 5\sqrt{3}$
E
$30$

Answer

The distance between object

A

and object

B

10\sqrt{3}

feet.

The horizontal distances from the base of the wall to objects

A

and

B

are calculated using the trigonometric ratios of the

30^\circ

60^\circ

90^\circ

and

45^\circ

45^\circ

90^\circ

triangles formed by the vertical wall. This gives legs of

5\sqrt{3}

feet and

15

feet. Applying the Pythagorean theorem to the right triangle on the ground yields a hypotenuse of

10\sqrt{3}

feet.

Step-by-Step Solution

Find the horizontal distance from the base of the wall to object

A

Let

O

be the base of the wall directly below the camera

C

, so

OC = 15

feet. The line of sight

CA

makes a

30^\circ

angle with the wall, so

\triangle OCA

is a

30^\circ

60^\circ

90^\circ

right triangle with leg

OA = 15 \tan(30^\circ) = \frac{15}{\sqrt{3}} = 5\sqrt{3}

feet.

We need to determine the length of one of the perpendicular legs on the ground.

Find the horizontal distance from the base of the wall to object

B

The line of sight

CB

makes a

45^\circ

angle with the wall, so

\triangle OCB

is a

45^\circ

45^\circ

90^\circ

right triangle with leg

OB = 15 \tan(45^\circ) = 15

feet.

We need to determine the length of the other perpendicular leg on the ground.

Use the Pythagorean theorem to calculate the straight-line distance

AB

on the ground.

Since the paths

OA

and

OB

are perpendicular,

\triangle AOB

is a right triangle with legs

OA = 5\sqrt{3}

and

OB = 15

. The hypotenuse

AB

\sqrt{(5\sqrt{3})^2 + 15^2} = \sqrt{75 + 225} = \sqrt{300} = 10\sqrt{3}

feet.

The straight-line distance between the two objects corresponds to the hypotenuse of the right triangle formed by their ground distances.

Key Concept

Applying special right triangle ratios and the Pythagorean theorem to solve multi-step problems in three-dimensional contexts.

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