Angle of Depression
The Angle of Depression is a key concept in CBSE Class 10 Mathematics, Chapter 9 (Some Applications of Trigonometry). It arises when an observer at a higher position looks downward at an object below.
The angle of depression is the angle between the horizontal line of sight (at the observer's eye level) and the line of sight directed downward to the object. It is the complement counterpart of the angle of elevation.
A crucial property: the angle of depression from point A (higher) to point B (lower) equals the angle of elevation from point B to point A. This follows from the alternate interior angles theorem when a horizontal line and the line of sight act as parallel lines cut by a transversal.
What is Angle of Depression - Definition, Formula & Solved Examples?
Definition: The angle of depression of an object as seen by an observer situated at a higher point is the angle formed between the horizontal line of sight and the line of sight directed downward to the object.
Key elements:
- Observer — positioned at a higher point (top of a building, cliff, lighthouse, etc.).
- Object — situated below the horizontal level of the observer.
- Horizontal line — imaginary line from the observer's eye, parallel to the ground.
- Line of sight — straight line from the observer's eye to the object.
- Angle of depression (θ) — angle between the horizontal and the downward line of sight.
Critical Property:
Angle of Depression from A to B = Angle of Elevation from B to A
This is because the horizontal at A and the ground at B are parallel lines, and the line of sight is a transversal. The two angles are alternate interior angles, hence equal.
Angle of Depression Formula
Formula for Angle of Depression Problems:
Since the angle of depression equals the angle of elevation (alternate interior angles), the same right-triangle trigonometric ratios apply:
tan θ = Perpendicular / Base = Height of Observer above Object / Horizontal Distance
Where:
- θ = angle of depression (= angle of elevation at the object)
- Height = vertical distance between the observer and the object
- Horizontal Distance = distance along the ground between the points directly below the observer and the object
Other useful forms:
- sin θ = Height / Line of Sight
- cos θ = Horizontal Distance / Line of Sight
To find height: h = d × tan θ
To find distance: d = h / tan θ = h × cot θ
Derivation and Proof
Establishing the Right Triangle:
- Let the observer be at point A, at height h above the ground (e.g., top of a building).
- Let the object be at point C on the ground.
- Let B be the point on the ground directly below A. So AB = h (height), BC = d (horizontal distance).
- The horizontal at A is parallel to the ground BC.
- ∠DAC = angle of depression = θ (where D is along the horizontal from A).
Using Alternate Interior Angles:
- AD ∥ BC (both horizontal).
- AC is a transversal cutting these parallel lines.
- ∠DAC (angle of depression) = ∠ACB (angle of elevation) — alternate interior angles.
In right triangle ABC (∠B = 90°):
- tan θ = AB/BC = h/d
- This is the same equation as for angle of elevation.
Key Takeaway: To solve angle-of-depression problems, mark the angle of depression at the top, transfer it to the base as the angle of elevation (using alternate interior angles), and then apply tan θ = h/d in the right triangle.
Types and Properties
Problems on angle of depression in Class 10 examinations include:
Type 1: Finding Distance from Height and Angle
- Given: observer's height and angle of depression.
- Find: horizontal distance to the object.
- Use: d = h/tan θ.
Type 2: Finding Height from Distance and Angle
- Given: horizontal distance and angle of depression.
- Find: height of the observer above the object.
- Use: h = d × tan θ.
Type 3: Two Objects on the Same Side
- From the top of a building/lighthouse, angles of depression to two objects on the same side are given.
- Form two right triangles and solve simultaneously.
Type 4: Two Objects on Opposite Sides
- Objects are on opposite sides of the building. Two angles of depression are given.
- Find the distance between the objects (sum of the two horizontal distances).
Type 5: Combined Elevation and Depression
- From a point, the angle of elevation to one object and the angle of depression to another are given.
- Form two triangles and solve.
Type 6: Moving Object (Boat/Car)
- From a fixed point (lighthouse/cliff), the angle of depression to a moving object changes as it approaches or recedes.
Methods
Step-by-step method for solving angle of depression problems:
- Draw the figure — place the observer at the top, the object below, and mark the horizontal line at the observer's level.
- Mark the angle of depression — between the horizontal and the downward line of sight.
- Transfer the angle — mark the equal angle of elevation at the object's position (alternate interior angles).
- Identify the right triangle — with the vertical height, horizontal distance, and line of sight as sides.
- Choose the ratio — tan θ (most common), sin θ, or cos θ based on what is given and required.
- Solve — substitute known values and solve for the unknown.
Common Mistakes to Avoid:
- Placing the angle of depression at the wrong vertex. It is at the observer's position, NOT at the object.
- Forgetting to use alternate interior angles to transfer the angle to the right triangle.
- Confusing angle of depression with the angle inside the right triangle at the top vertex. The angle of depression = 90° minus the angle at the top vertex of the triangle.
- Not accounting for the height of the observer above the ground when the observer is standing on a building.
Solved Examples
Example 1: Finding Distance to a Boat
Problem: From the top of a lighthouse 75 m above sea level, the angle of depression of a boat is 30°. Find the distance of the boat from the base of the lighthouse.
Solution:
Given:
- Height of lighthouse (h) = 75 m
- Angle of depression (θ) = 30°
Using tan θ = h/d:
- tan 30° = 75/d
- 1/√3 = 75/d
- d = 75√3
- d = 75 × 1.732 = 129.9 m
Answer: The boat is 75√3 m ≈ 129.9 m from the base of the lighthouse.
Example 2: Finding Height of a Cliff
Problem: From the top of a cliff, the angle of depression of a car parked on the ground is 60°. If the car is 50 m from the base of the cliff, find the height of the cliff.
Solution:
Given:
- Horizontal distance (d) = 50 m
- Angle of depression (θ) = 60°
Using tan θ = h/d:
- tan 60° = h/50
- √3 = h/50
- h = 50√3
- h = 50 × 1.732 = 86.6 m
Answer: The height of the cliff is 50√3 m ≈ 86.6 m.
Example 3: Two Objects on the Same Side
Problem: From the top of a 100 m building, the angles of depression of two cars on the same side are 30° and 45°. Find the distance between the two cars.
Solution:
Given:
- Height (h) = 100 m
- Angle to nearer car = 45°, Angle to farther car = 30°
Distance to nearer car (d₁):
- tan 45° = 100/d₁ → 1 = 100/d₁ → d₁ = 100 m
Distance to farther car (d₂):
- tan 30° = 100/d₂ → 1/√3 = 100/d₂ → d₂ = 100√3 m
Distance between cars:
- = d₂ − d₁ = 100√3 − 100 = 100(√3 − 1)
- = 100(1.732 − 1) = 100 × 0.732 = 73.2 m
Answer: The distance between the cars is 100(√3 − 1) m ≈ 73.2 m.
Example 4: Two Objects on Opposite Sides
Problem: From the top of a 60 m tower, the angles of depression of two points on opposite sides of the tower are 30° and 60°. Find the distance between the two points.
Solution:
Given:
- Height (h) = 60 m
- Angles of depression: 30° and 60° (on opposite sides)
Distance to point A (angle 30°):
- tan 30° = 60/d₁ → 1/√3 = 60/d₁ → d₁ = 60√3 m
Distance to point B (angle 60°):
- tan 60° = 60/d₂ → √3 = 60/d₂ → d₂ = 60/√3 = 20√3 m
Total distance (opposite sides, so add):
- = d₁ + d₂ = 60√3 + 20√3 = 80√3
- = 80 × 1.732 = 138.56 m
Answer: The distance between the two points is 80√3 m ≈ 138.56 m.
Example 5: Moving Boat Problem
Problem: From a cliff 80 m high, the angle of depression of a boat changes from 30° to 60° as it approaches the cliff. Find the distance travelled by the boat.
Solution:
Given:
- Height of cliff (h) = 80 m
- Initial angle of depression = 30°
- Final angle of depression = 60°
Initial distance (d₁):
- tan 30° = 80/d₁ → d₁ = 80√3 m
Final distance (d₂):
- tan 60° = 80/d₂ → d₂ = 80/√3 = 80√3/3 m
Distance travelled:
- = d₁ − d₂ = 80√3 − 80√3/3
- = 80√3(1 − 1/3) = 80√3 × 2/3
- = 160√3/3 ≈ 92.38 m
Answer: The boat travelled 160√3/3 m ≈ 92.38 m.
Example 6: Finding Line of Sight Length
Problem: From the top of a 40 m tower, the angle of depression of a point on the ground is 45°. Find the distance of that point from the top of the tower (line of sight).
Solution:
Given:
- Height (h) = 40 m
- Angle of depression (θ) = 45°
Using sin θ = h/L:
- sin 45° = 40/L
- 1/√2 = 40/L
- L = 40√2
- L = 40 × 1.414 = 56.56 m
Answer: The line of sight is 40√2 m ≈ 56.57 m.
Example 7: Balloon Problem
Problem: A balloon is at a height of 50 m. A person standing on the ground observes the angle of elevation as 30°. After some time, the angle of depression of the same person from the balloon is 60° when viewed from the balloon. Verify that both angles give the same horizontal distance.
Solution:
Given:
- Height (h) = 50 m
- Angle of elevation from ground = 30°
- Angle of depression from balloon = 60°
From ground (angle of elevation 30°):
- tan 30° = 50/d → 1/√3 = 50/d → d = 50√3 m
From balloon (angle of depression 60°):
- Angle of depression from balloon = angle of elevation from ground? No — the problem states two different angles.
- tan 60° = 50/d₂ → √3 = 50/d₂ → d₂ = 50/√3 = 50√3/3 m
Since 50√3 ≠ 50√3/3, the two distances are different. This means the balloon has moved horizontally. The person's position changed relative to the balloon.
Answer: The horizontal distances are 50√3 m and 50√3/3 m respectively — different because the problem describes two different moments.
Example 8: Bird on a Tree
Problem: A bird sitting on a tree 12 m high is observed from a window at a height of 4 m at an angle of elevation of 60°. Find the distance of the tree from the building.
Solution:
Given:
- Height of bird = 12 m
- Height of window = 4 m
- Angle of elevation to bird = 60°
Effective height (above window level):
- h = 12 − 4 = 8 m
Using tan θ = h/d:
- tan 60° = 8/d
- √3 = 8/d
- d = 8/√3 = 8√3/3
- d ≈ 4.62 m
Answer: The distance of the tree from the building is 8√3/3 m ≈ 4.62 m.
Real-World Applications
Maritime Navigation:
- Lighthouse keepers and coast guard personnel measure angles of depression to ships to calculate their distance from shore.
Aviation:
- Pilots use the angle of depression to the runway during landing approach to maintain the correct glide slope (typically 3°).
Military:
- Artillery and sniper calculations use angles of depression to targets below the firing position for accurate aiming.
Surveying:
- Surveyors measure angles of depression from hilltops to determine elevations and distances of points in valleys.
Construction:
- Engineers calculate the slope of drainage pipes and ramps using angles of depression to ensure proper gradient.
Rescue Operations:
- Helicopter rescue teams calculate the angle of depression to a stranded person to determine rope length and approach path.
Key Points to Remember
- The angle of depression is the angle between the horizontal line of sight and the downward line of sight to an object below.
- It is measured at the observer's position (at the higher point), NOT at the object.
- The angle of depression from A to B equals the angle of elevation from B to A (alternate interior angles).
- To solve problems, transfer the angle of depression to the angle of elevation at the base using alternate interior angles.
- The most used ratio is tan θ = height/distance.
- As the object moves farther away, the angle of depression decreases.
- As the object moves closer, the angle of depression increases.
- Both the angle of elevation and the angle of depression are always between 0° and 90°.
- In combined problems (both elevation and depression from the same point), two separate right triangles are formed.
- Always draw a clear, labelled figure before solving.
Practice Problems
- From the top of a 120 m cliff, the angle of depression of a boat at sea is 30°. Find the distance of the boat from the cliff's base.
- From the top of a building 50 m high, the angle of depression of a car is 60°. Find the distance of the car from the building.
- From a helicopter flying at 300 m, the angles of depression of two ships on opposite sides are 45° and 60°. Find the distance between the ships.
- From a window 15 m above the ground, the angle of depression of the foot of a wall across the street is 30°. Find the width of the street.
- A man on top of a 30 m tower observes a car approaching. The angle of depression changes from 30° to 45°. Find the distance the car has moved.
- From the top of a 200 m hill, the angles of depression of the top and bottom of a tower are 30° and 60° respectively. Find the height of the tower.
Frequently Asked Questions
Q1. What is an angle of depression?
The angle of depression is the angle between the horizontal line of sight and the line of sight directed downward to an object below the observer's level. It is measured at the observer's position.
Q2. Is the angle of depression equal to the angle of elevation?
Yes. The angle of depression from point A (higher) to point B (lower) equals the angle of elevation from B to A. This is because the horizontal at A is parallel to the ground at B, and the line of sight is a transversal, making them alternate interior angles.
Q3. Where is the angle of depression measured?
The angle of depression is measured at the observer's position (the higher point). It is the angle between the observer's horizontal line of sight and the downward line of sight to the object.
Q4. How do you solve angle of depression problems?
Draw the figure, mark the angle of depression at the observer's position. Use the alternate interior angles property to transfer it to the right triangle as an angle of elevation. Then apply tan θ = height/distance (or sin θ, cos θ as needed) to solve.
Q5. Can the angle of depression be greater than 90°?
No. The angle of depression is between the horizontal and the downward line of sight, so it ranges from 0° (looking straight ahead) to just under 90° (looking almost straight down).
Q6. What is the difference between angle of depression and angle of inclination?
Angle of depression is measured downward from the horizontal. Angle of inclination (or elevation) is measured upward from the horizontal. They describe opposite directions of looking.
Q7. Why do we use alternate interior angles in depression problems?
The horizontal at the observer's level is parallel to the ground. The line of sight acts as a transversal. The angle of depression (above) and the angle of elevation (below) are alternate interior angles formed by this transversal cutting two parallel lines.
Q8. What happens when two angles of depression are given from the same point?
Each angle forms a separate right triangle with the same height but different horizontal distances. Solve each triangle using tan θ = h/d. The distance between the objects is the difference (same side) or sum (opposite sides) of these horizontal distances.
