Two friends, each with eyes 1.5 m above the ground, stand on level ground on the same side of a vertical tower and in line with its foot, 20 m apart. The nearer one sees the top of the tower at an elevation of 60° and the farther one at 30°. The height of the tower is:

Both elevations are measured from eye level, so work first with h, the part of the tower ABOVE eye level. The two standpoints are h·cot60° and h·cot30° from the foot, and their difference is the 20 m gap: h(√3 − 1/√3) = 20, i.e. h × 2/√3 = 20, so h = 10√3 ≈ 17.32 m (the stations are 10 m and 30 m out). The tower is measured from the ground, so the eye height must be put back: 10√3 + 1.5 ≈ 18.82 m. Stopping at h gives 10√3 — the classic slip of forgetting the observer's own height — while adding both friends' heights gives 10√3 + 3, and treating 20 m as the near distance gives 20√3 + 1.5.