Angles of Depression in 3D: Helicopter and Fire Truck Problem

Grades 10-11

Interactive Angles of Depression in 3D: Helicopter and Fire Truck Problem graph — click to explore — Click the graph to explore it interactively →

Question: Please turn this problem into a GGB file, embedded in a rectangular box. I need the exact values from the problem, not approximations, keep them fixed, and add icons for the fire helicopter, the fire location, and the fire truck.

In this problem, we need to find the ground distance between two points,

A

(the fire location) and

B

(the fire truck). This is a classic 3D geometry problem, which we can make easier to understand by embedding it inside a rectangular box model.

Mathematical Setup and Derivation

We set up a 3D coordinate system, taking the helicopter's projection onto the ground as the origin

O(0, 0, 0)

The Helicopter's Position: The helicopter sits $300$ metres directly above $O$ , at $H(0, 0, 300)$ .
The Fire Location $A$ :
- $A$ lies due west of the helicopter, with an angle of depression of $60^\circ$ . This means the angle of elevation from $A$ up to the helicopter $H$ is also $60^\circ$ .
- In right triangle $\triangle AOH$ , $\angle HOA = 90^\circ$ and the height $OH = 300$ .
- By trigonometry: $\tan(60^\circ) = \frac{OH}{OA}$ , so $OA = \frac{300}{\tan(60^\circ)} = \frac{300}{\sqrt{3}} = 100\sqrt{3}$ metres.
- Taking west as the negative $x$ -direction, $A$ 's coordinates are $(-100\sqrt{3}, 0, 0)$ .
The Fire Truck's Position $B$ :
- $B$ lies due south of the helicopter, with an angle of depression of $30^\circ$ . Likewise, the angle of elevation from $B$ up to $H$ is $30^\circ$ .
- In right triangle $\triangle BOH$ , $\angle HOB = 90^\circ$ and the height $OH = 300$ .
- By trigonometry: $\tan(30^\circ) = \frac{OH}{OB}$ , so $OB = \frac{300}{\tan(30^\circ)} = \frac{300}{1/\sqrt{3}} = 300\sqrt{3}$ metres.
- Taking south as the negative $y$ -direction, $B$ 's coordinates are $(0, -300\sqrt{3}, 0)$ .
Finding the Distance Between $A$ and $B$ :
- Since due west and due south are perpendicular directions, $\triangle AOB$ is a right triangle with $\angle AOB = 90^\circ$ .
- By the Pythagorean theorem: $\begin{align*} AB^2 &= OA^2 + OB^2 \\ AB^2 &= (100\sqrt{3})^2 + (300\sqrt{3})^2 \\ AB^2 &= 30{,}000 + 270{,}000 \\ AB^2 &= 300{,}000 \\ AB &= \sqrt{300{,}000} = 100\sqrt{30} \approx 547.72 \text{ metres} \end{align*}$

About the Diagram

In the interactive figure below:

Blue rectangular box: represents the spatial boundary of the scene, helping us locate $A$ , $B$ , and $H$ .
Red point (H): the fire helicopter.
Orange point (A): the fire location.
Blue point (B): the fire truck.
Grey triangles: show the vertical angle-of-elevation relationships ( $60^\circ$ and $30^\circ$ ) and the right angle on the ground.

Drag the rotation slider to view this 3D structure from different angles.