Elementary heights and distances

Learning outcome

Model heights and horizontal distances using right triangles, include observer eye height, and solve a two-position observation problem.

Concepts and assumptions

All angles are in degrees with 0° < θ < 90°. Assume level horizontal ground, vertical targets and straight sight lines; ignore Earth curvature and refraction. Treat dimensions as ideal values.

Elevation is measured upward from the observer’s horizontal eye line; depression is measured downward from it.

Let tower BT have height H and base B. An observer stands at P on the same ground level, with eyes E at height e above P. Set PB = d > 0. If T is above E, a horizontal line from E meets the tower at R; ER = d and RT = H − e.

Thus tan θ = (H − e)/d, or H = e + d tan θ. The sight line ET is the hypotenuse, not horizontal distance.

Parallel horizontal lines make a depression angle equal to the corresponding elevation angle. Round only specified final answers.

Worked examples

Example 1 — Eye height matters. An observer’s eyes are 1.5 m above level ground, 18 m horizontally from a tree’s base. Elevation to the top is 45°.

Height above the eyes = 18 tan 45° = 18 m. Tree height = 1.5 + 18 = 19.5 m.

Example 2 — Depression. Eyes are 24 m above level ground. Depression to a ground marker is 30°. Find horizontal distance from the point directly below the eyes.

The corresponding elevation is 30°, so tan 30° = 24/d. Thus d = 24/(1/√3) = 24√3 m, exactly.

Example 3 — Two positions. Stations P and Q are 20 m apart on one straight level line, on the same side of tower base B. Eyes are 1.6 m high; elevations are 60° at nearer P and 30° at farther Q.

Set PB = d and h = H − 1.6; then QB = d + 20. h = d√3 and h = (d + 20)/√3. Equating gives 3d = d + 20, hence d = 10 m. Therefore H = 1.6 + 10√3 m ≈ 18.92 m, rounded finally to two decimal places.

Common mistakes

Omitting eye height; confusing sloping and horizontal distances; measuring from the vertical; using d + gap without checking station positions.

Practice questions

1. On level ground, eyes are 1.4 m high, 28 m horizontally from a tower’s base. Elevation is 45°. Find tower height.
2. A tower is 25 m high. On the same level ground, eyes are 1 m high and elevation is 60°. Find horizontal distance.
3. Eyes are 30 m above level ground; depression to a ground marker is 45°. Find horizontal distance from the point directly below the eyes.
4. Two stations on the same straight level line and same side of a tower are 12 m apart. Eye height is 1.5 m at both; nearer elevation is 45°, farther elevation 30°. Find exact tower height.

Worked solutions

1. Height above eyes = 28 tan 45° = 28 m. Total = 28 + 1.4 = 29.4 m.
2. Vertical difference = 25 − 1 = 24 m. Distance = 24/tan 60° = 24/√3 = 8√3 m.
3. tan 45° = 30/d, so 1 = 30/d and d = 30 m.
4. Let nearer distance be d and h = H − 1.5. Then h = d and h = (d + 12)/√3. Thus (√3 − 1)d = 12, giving d = 12/(√3 − 1) = 6(√3 + 1). Therefore H = 7.5 + 6√3 m, exactly.