Angles of Elevation and Depression
Model real-world problems involving angles of elevation and depression with right triangle diagrams. Apply trigonometric ratios to find heights of buildings, depths of canyons, and other indirect measurements.
A lesson of Geometry, in Right Triangles and Trigonometry, about 10 minutes.
By the end of this lesson, you will be able to define angles of elevation and depression, draw accurate diagrams for real-world problems, and solve for heights and distances using trigonometric ratios.
The Big Idea
Angles of elevation and depression are the practical faces of trigonometry. Surveyors use them to measure mountain heights; pilots use them to calibrate glide paths; architects use them for sight lines. Every problem reduces to identifying a right triangle in the scenario and applying the correct trig ratio.
Definitions
Angle of elevation: the angle formed between the horizontal and the line of sight when looking upward toward an object.
Angle of depression: the angle formed between the horizontal and the line of sight when looking downward toward an object.
Key fact: The angle of elevation from point $A$ to point $B$ equals the angle of depression from point $B$ to point $A$ — they are alternate interior angles with a horizontal transversal.
Setting Up the Diagram
- Draw the horizontal reference line at the observer's eye level
- Draw the line of sight
- Mark the angle (elevation = above horizontal; depression = below)
- Identify the right triangle (horizontal, vertical, line of sight)
- Label the known and unknown sides/angles
Example 1 — Height of a Building
From a point 80 feet from the base of a building, the angle of elevation to the top is 55°. Find the height of the building.
The horizontal distance is the adjacent leg, the building height is the opposite leg.
$$\tan 55° = \frac{h}{80} \implies h = 80\tan 55° \approx 80(1.4281) \approx 114.2 \text{ feet}$$
Example 2 — Angle of Depression
A lifeguard sits in a tower 15 feet above the ground. They spot a swimmer at an angle of depression of 18°. How far is the swimmer from the base of the tower?
The angle of depression = angle of elevation from swimmer to lifeguard = 18°.
The height is the opposite leg (15 ft); the horizontal distance is the adjacent leg.
$$\tan 18° = \frac{15}{d} \implies d = \frac{15}{\tan 18°} \approx \frac{15}{0.3249} \approx 46.2 \text{ feet}$$