358. Chasing the Cloud's Height · using angles of elevation and depression
The angles of elevation and depression match the problem's 30° and 60° when the horizontal distance is correct.
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For subscribers · Selina ICSE: Heights and Distances
What this lesson covers
Try to break it
Drag E to shift the cloud's vertical alignment over the ground. The angle of elevation up to the cloud and the angle of depression down to its reflection both change with E. The cloud's height h is recovered from h = d · tan(elevation), where d is the horizontal distance from the observer to E.
The proof, step by step
Prove that the height of the cloud follows from the 30° and 60° angles.
- In right ΔOEC, tan(30°) = CE / OE = (h − 25) / d. Thus, d = √3(h − 25).
- In right ΔOER, tan(60°) = ER / OE = (h + 25) / d. Thus, d = (h + 25) / √3.
- Equating both expressions for d: √3(h − 25) = (h + 25) / √3.
- Simplifying gives 3(h − 25) = h + 25 → 2h = 100 → h = 50 m.
Worked example
The angle of elevation of a cloud from a point 25 m above a lake is 30° and the angle of depression of its reflection is 60°. Find the height of the cloud above the lake level.
Form two right triangles sharing horizontal distance d. tan 30° = (h-25)/d and tan 60° = (h+25)/d. Equating d gives √3(h-25) = (h+25)/√3. Solving 3h-75=h+25 yields 2h=100, so h=50 m.
- 35 m
- 45 m
- 50 m — correct
- 60 m