3.3 Navigation

1. Angle of Elevation and Angle of Depression

Angles measured relative to a horizontal line define positions in space.

Definitions:

  • Angle of Elevation: The upward angle from an observer's horizontal sightline to an object above them.
  • Angle of Depression: The downward angle from an observer's horizontal sightline to an object 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 $4\text{m}$, depth of $5\text{m}$, and height of $3\text{m}$. An observer is at corner $A$. Calculate the angles to the opposite corners.

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 right triangle ($ABF$) with base $AB = 4\text{m}$ and height $BF = 3\text{m}$. The angle of elevation from $A$ to $F$ uses the tangent ratio:
$$\tan(\theta) = \dfrac{3}{4} \implies \theta = \mathbf{36.9^\circ}$$
(b) Angle of Elevation of the opposite space corner $G$:
First, calculate the base diagonal $AC$ using the Pythagorean theorem:
$$AC^2 = 4^2 + 5^2 \implies AC = \sqrt{16 + 25} = \sqrt{41}\text{m}$$
The space diagonal $AG$ has a vertical height $CG = 3\text{m}$ and a 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 $P$ is vertically above point $Q$. Observers $A$ and $B$ are $10\text{m}$ apart along a straight horizontal line. The angle of elevation from $A$ is $45^\circ$ and from $B$ is $30^\circ$. Calculate the height $h$.

B A P Q 30° 45° 10m x h
Let the horizontal distance from station $A$ to $Q$ be $x$.
Triangle $PAQ$ equation:
$$\tan 45^\circ = \dfrac{h}{x} \implies 1 = \dfrac{h}{x} \implies h = x$$
Triangle $PBQ$ equation:
$$\tan 30^\circ = \dfrac{h}{x+10} \implies \dfrac{1}{\sqrt{3}} = \dfrac{h}{h+10}$$
Solve for $h$:
$$\begin{aligned} h\sqrt{3} &= h + 10 \\ h\sqrt{3} - h &= 10 \\ h(\sqrt{3} - 1) &= 10 \\ h &= \dfrac{10}{\sqrt{3}-1} \approx \mathbf{13.7\text{m}} \end{aligned}$$

2. Navigation and Bearing

Navigation uses angles called bearings.

  • A bearing indicates the direction of travel measured clockwise from North, written as a 3-digit number (e.g., $000^\circ$).
  • Standard directions: North ($000^\circ$), East ($090^\circ$), South ($180^\circ$), West ($270^\circ$).
  • Intermediate directions: Northeast is $045^\circ$ and Southwest is $225^\circ$.

Application: If the path from $A$ to $B$ has a bearing of $050^\circ$, the return path from $B$ to $A$ changes by $180^\circ$, giving a bearing of $230^\circ$.

EXAMPLE 3

A vehicle travels from $A$ to $B$ on a bearing of $050^\circ$, then from $B$ to $C$ on a bearing of $150^\circ$, and returns directly to $A$ on a bearing of $270^\circ$. If $AC = 10\text{km}$, find distances $AB$ and $BC$.

A B C 10km 050° 40° 60° 80° c (AB) a (BC)
1. Find the interior angles of the triangle using parallel lines:
Since the path from $C$ to $A$ is due West ($270^\circ$), line $AC$ is horizontal.
  • $\angle A = 90^\circ - 50^\circ = \mathbf{40^\circ}$
  • The return bearing to $A$ is $50^\circ + 180^\circ = 230^\circ$. The path from $B$ to $C$ is $150^\circ$.
    $\angle B = 230^\circ - 150^\circ = \mathbf{80^\circ}$
  • The return bearing to $B$ is $150^\circ + 180^\circ = 330^\circ$. The path from $C$ to $A$ is $270^\circ$.
    $\angle C = 330^\circ - 270^\circ = \mathbf{60^\circ}$
Check: $40^\circ + 80^\circ + 60^\circ = 180^\circ$.
2. Apply the Law of Sines:
$$\dfrac{10}{\sin 80^\circ} = \dfrac{AB}{\sin 60^\circ} = \dfrac{BC}{\sin 40^\circ}$$
3. Solve for the lengths:
$$\begin{aligned} AB &= \dfrac{10 \sin 60^\circ}{\sin 80^\circ} \approx \mathbf{8.79\text{km}} \\ BC &= \dfrac{10 \sin 40^\circ}{\sin 80^\circ} \approx \mathbf{6.53\text{km}} \end{aligned}$$