← Unit 4 Angles of Elevation and Depression

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Angles of Elevation and Depression

⏱ 25 min📚 Year 9📈
Think First

You stand 30 m from a building and look up at the top at $40°$. How tall is the building?

💡 Revision: Ensure you can solve right-angled triangles using SOH CAH TOA.

Learning Intentions

Know

  • Angle of elevation
  • Angle of depression
  • Horizontal line
  • Line of sight

Understand

  • Why elevation and depression angles are equal
  • How to draw an accurate diagram

Can Do

  • Identify elevation and depression in a problem
  • Draw a labelled diagram
  • Calculate unknown heights or distances
ElevationDepressionLine of sightHorizontal
Learn Phase
1

Angle of Elevation

Looking up from the horizontal

The angle of elevation is the angle between the horizontal and the line of sight when looking up at an object.

Key Property

If the observer and object are at different heights, form a right triangle with the horizontal distance.

Example: Standing 20 m from a tower, looking up at $30°$. Height above eye level:

$ an 30° = h/20$ → $h = 20 imes an 30° approx 11.55$ m

2

Angle of Depression

Looking down from the horizontal

The angle of depression is the angle between the horizontal and the line of sight when looking down from an elevated position.

Crucially: angle of elevation = angle of depression

This is because they are alternate angles formed by parallel horizontal lines.

3

Drawing Diagrams

The key to solving these problems

Always draw a clear diagram:

  1. Mark the observer and the object
  2. Draw horizontal lines through each
  3. Mark the angle of elevation/depression
  4. Label known distances and heights
  5. Identify the right triangle

Tip: Add the observer's eye height to calculated heights for total height above ground.

✓

Check Understanding

Try it yourself

From a cliff 50 m high, the angle of depression to a boat is $20°$. How far is the boat from the cliff base?

Worked Example

Elevation and Depression

1

A person 1.7 m tall stands 30 m from a tree and looks up at the top at $35°$. Find the tree height.

$ an 35° = h/30$ → $h = 30 imes an 35° approx 21.0$ m

Total height = $21.0 + 1.7 = 22.7$ m

2

From a tower 40 m high, the angle of depression to a car is $25°$. How far is the car?

$ an 25° = 40/d$ → $d = 40/ an 25° approx 85.8$ m

3

A plane at 3000 m sees a runway at $3°$ depression. How far horizontally?

$ an 3° = 3000/d$ → $d = 3000/ an 3° approx 57,300$ m $= 57.3$ km

Common Misconceptions

✕

Angle of elevation and depression are different. They are equal (alternate angles). Do not calculate them differently.

✕

Forget to add the observer's height. If the observer is 1.7 m tall and calculates height above eye level as 20 m, the total is 21.7 m.

✕

Use the wrong trig ratio. Elevation/depression problems typically use tangent because you have horizontal distance and vertical height.

Your Turn

Practice — Elevation and Depression

Work through each question in your book or digitally. Answers are in the Questions phase.

1A 30 m cliff, angle of depression $25°$ to a swimmer. How far from the cliff?
2A flagpole casts a 15 m shadow. The sun's angle of elevation is $50°$. Find the flagpole height.
3From a plane at 5000 m, angle of depression to an airport is $2°$. Horizontal distance?
Real-World Anchor

Aviation and Maritime

Pilots use angles of depression to calculate distance to landing strips. Maritime navigation uses these principles to determine distance to shore. The Sydney Harbour Bridge was designed using these calculations to ensure proper clearance heights.

📓 Copy Into Your Books

▼

Elevation

  • Angle up from horizontal
  • Use $ an = ext{opp}/ ext{adj}$

Depression

  • Angle down from horizontal
  • Equals angle of elevation from below

Diagram

  • Draw horizontal lines
  • Mark parallel lines
  • Label all knowns
Questions Phase
Check Your Understanding
Answer all questions correctly to unlock the Game phase.
Angle of elevation is measured from:
Angle of elevation = angle of depression because:
Standing 20 m away, elevation $30°$. Height above eye level:
From 50 m cliff, depression $20°$ to boat. Distance:
Elevation problems usually use:
A 1.6 m person sees top of 10 m tree at elevation $ heta$. $ an heta = $
From 100 m building, depression $10°$ to car. Distance:
Shadow 12 m, sun elevation $40°$. Object height:
1From a 40 m cliff, depression to a ship is $15°$. Find the ship's distance.
2A 1.5 m observer 25 m from a statue looks up at $50°$. Find the statue height.
3Explain why angles of elevation and depression are equal.

Comprehensive Answers

140 m cliff, depression $15°$.
$d = 40/ an 15° approx 149$ m.
21.5 m observer, 25 m away, $50°$ elevation.
$h = 25 imes an 50° + 1.5 approx 31.3$ m.
3Why elevation = depression.
Alternate angles between parallel horizontal lines.
MC 1Elevation measured from.
Horizontal. Answer: B
MC 2Why equal.
Alternate angles. Answer: A
MC 320 m away, $30°$ elevation.
$20 imes an 30° approx 11.5$ m. Answer: B
MC 450 m cliff, $20°$ depression.
$50/ an 20° approx 137$ m. Answer: C
MC 5Elevation problems use.
Tangent. Answer: C
MC 6$ an heta$ for tree.
$(10-1.6)/d = 8.4/d$. Answer: B
MC 7100 m, $10°$ depression.
$100/ an 10° approx 567$ m. Answer: C
MC 8Shadow 12 m, $40°$ elevation.
$12 imes an 40° approx 10.1$ m. Answer: B
SA 140 m cliff, $15°$ depression.
$40/ an 15° approx 149$ m.
SA 2Statue height.
$25 imes an 50° + 1.5 approx 31.3$ m.
SA 3Why equal angles.
Alternate angles between parallel horizontal lines.
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