3.3 Applications in 3D Geometry & Navigation

1. Angle of Elevation and Angle of Depression

Angles measured relative to horizontal perspectives define spatial relationships critically in applied environments.

Definitions:

  • Angle of Elevation: The upward angle measured from an observer's direct horizontal sightline to an object physically situated above them.
  • Angle of Depression: The downward angle measured from an observer's horizontal sightline to an object situated below them.
Angle of Elevation Angle of Depression Object Object Observer

EXAMPLE 1 (3D Cuboid Evaluation)

Consider a rectangular cuboid with a base width of $4\text{m}$, a base depth of $5\text{m}$, and a vertical height of $3\text{m}$. An observer rests at the bottom corner point $A$. Calculate the angles referencing the opposite structure points.

A B C D E F G H 4m 5m 3m
(a) Angle of Elevation of point $F$ (across the front face):
The front face forms a standard 2D right triangle ($ABF$) with base $AB = 4\text{m}$ and height $BF = 3\text{m}$. The angle of elevation from $A$ to $F$ utilizes the tangent ratio relative to the height:
$$\tan(\theta) = \dfrac{3}{4} \implies \theta = \mathbf{36.9^\circ}$$
(b) Angle of Elevation of the extreme opposite space corner $G$:
First, evaluate the flat diagonal base $AC$ strictly using Pythagoras' theorem on the base dimensions ($4\text{m}$ and $5\text{m}$):
$$AC^2 = 4^2 + 5^2 \implies AC = \sqrt{16 + 25} = \sqrt{41}\text{m}$$
The true 3D space diagonal $AG$ requires tracking the vertical height $CG = 3\text{m}$ rising strictly over the flat base diagonal projection $AC = \sqrt{41}\text{m}$:
$$\tan(\phi) = \dfrac{3}{\sqrt{41}} \implies \phi \approx \mathbf{25.1^\circ}$$

EXAMPLE 2 (Multiple Observer Scenario)

An object $P$ hovers vertically above a hill base. Observers $A$ and $B$ are situated $10\text{m}$ apart along a straight horizontal tracking line. The measured angle of elevation from point $A$ is $45^\circ$, and from point $B$ is $30^\circ$. Calculate the vertical height $h$.

B A P Q 30° 45° 10m x h
Set the interior horizontal distance from tracking station $A$ to the target base center as $x$.
Triangle $PAQ$ evaluates directly:
$$\tan 45^\circ = \dfrac{h}{x} \implies 1 = \dfrac{h}{x} \implies h = x$$
Triangle $PBQ$ evaluates compositely:
$$\tan 30^\circ = \dfrac{h}{x+10} \implies \dfrac{1}{\sqrt{3}} = \dfrac{h}{h+10}$$
Algebraic isolation of $h$:
$$\begin{aligned} h\sqrt{3} &= h + 10 \\ h\sqrt{3} - h &= 10 \\ h(\sqrt{3} - 1) &= 10 \\ h &= \dfrac{10}{\sqrt{3}-1} \approx \mathbf{13.7\text{m}} \end{aligned}$$

2. Navigation and Bearing

Geographic navigation utilizes absolute angular coordinates referred to universally as Bearings.

  • A bearing always indicates the exact direction of travel measured clockwise continuously from the absolute North direction (written as a 3-digit number like $000^\circ$).
  • Standard primary orientations: North ($000^\circ$), East ($090^\circ$), South ($180^\circ$), West ($270^\circ$).
  • Intermediate lines split evenly: Northeast evaluates to $045^\circ$, whereas Southwest equates to $225^\circ$.

Geometric Application: If path $A \to B$ holds a bearing of $050^\circ$, the reciprocal return path $B \to A$ shifts geometrically by $180^\circ$, creating a reciprocal bearing of $230^\circ$.

EXAMPLE 3

A vehicle executes a multipart survey route: From point $A$ to point $B$ heading at a bearing of $050^\circ$. Then, from point $B$ to $C$ heading at a bearing of $150^\circ$. Finally, it returns directly to $A$ heading at a bearing of $270^\circ$. The distance $AC$ evaluates to $10\text{km}$. Find the distances $AB$ and $BC$.

A B C 10km 050° 40° 60° 80° c (AB) a (BC)
1. Mapping the interior triangle angles utilizing parallel North/South lines determines:
Since $C \to A$ is exactly West ($270^\circ$), the path $AC$ rests perfectly horizontal relative to North.
  • $\angle A = 90^\circ - 50^\circ = \mathbf{40^\circ}$
  • Reciprocal bearing to $A$ is $50^\circ + 180^\circ = 230^\circ$. Path $B \to C$ is $150^\circ$.
    $\angle B = 230^\circ - 150^\circ = \mathbf{80^\circ}$
  • Reciprocal bearing to $B$ is $150^\circ + 180^\circ = 330^\circ$. Path $C \to A$ is $270^\circ$.
    $\angle C = 330^\circ - 270^\circ = \mathbf{60^\circ}$
Check: $40^\circ + 80^\circ + 60^\circ = 180^\circ$. The geometry holds correctly.
2. Implementing the Sine Rule across the full geometric sequence:
$$\dfrac{10}{\sin 80^\circ} = \dfrac{AB}{\sin 60^\circ} = \dfrac{BC}{\sin 40^\circ}$$
3. Algebraic evaluation determines the path lengths:
$$\begin{aligned} AB &= \dfrac{10 \sin 60^\circ}{\sin 80^\circ} \approx \mathbf{8.79\text{km}} \\ BC &= \dfrac{10 \sin 40^\circ}{\sin 80^\circ} \approx \mathbf{6.53\text{km}} \end{aligned}$$