The angle between the horizontal line and the upward line of sight when an observer looks at an object above eye level.
- Measured upward from horizontal
- Used in problems involving towers, buildings, trees
Chapter 9 · CBSE Class X
📘 Definition
Definition & Conceptual Foundation
📖 Theory
Concept
🎨 SVG Diagram
Angle of Elevation
🎨 SVG Diagram
Angle of Depression
The angle between the horizontal line and the downward line of sight when an observer looks at an object below eye level.
- Measured downward from horizontal
- Numerically equal to angle of elevation (alternate angles)
Line of Sight
An imaginary straight line joining the observer’s eye and the object. It forms the hypotenuse of the right triangle.
🔢 Formula
Important Formulae
\[ \begin{aligned}
\tan \theta &= \dfrac{\text{Height}}{\text{Distance}}\\\\
\text{Height} &= \text{Distance} \times \tan \theta\\\\
\text{Distance} &= \dfrac{\text{Height}}{\tan \theta}
\end{aligned} \]
✏️ Example
Illustrative Example (Board Level)
A tower stands on a horizontal ground. The angle of elevation of its top from a point on the ground is \(30^\circ\). If the point is 20 m away from the base, find the height of the tower.</p>
Concept Used: Right triangle + tangent ratio
\[
\begin{aligned}
\tan 30^\circ &= \dfrac{h}{20} \\
\frac{1}{\sqrt{3}} &= \dfrac{h}{20}\\
h &= \frac{20}{\sqrt{3}} \\&= \dfrac{20\sqrt{3}}{3} \text{ m}
\end{aligned}
\]
📐 Derivation
Derivation Insight
All height-distance problems reduce to a right triangle where:
- Height = Perpendicular
- Distance = Base
- Line of sight = Hypotenuse
Thus, using trigonometric definitions: \[ \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} \]
⚡ Exam Tip
Exam Tips (CBSE Focused)
⚠️ Warning
Common Mistakes
📋 Case Study
CBSE Case Study (HOTS)
A lighthouse is observed from a ship at sea. The angle of elevation of the top of the lighthouse is \(60^\circ\). After moving 40 m closer, the angle becomes \(75^\circ\). Find the height of the lighthouse.
🗺️ APPROACH ROADMAP- Let initial distance = x
- New distance = (x − 40)
- Apply tan θ in both cases
- Solve simultaneous equations
This type of question tests: concept clarity + equation formation + algebra
📄 IMPORTANCE
Why This Topic is Important for Board Exams
- Frequently asked in 2–4 mark questions
- Direct application of trigonometric ratios
- Scoring and less conceptual ambiguity
- Forms base for Class 11 applications
📘 Definition
Definition & Core Idea
🎨 SVG Diagram
Concept Visualisation (Triangle Mapping)
- Base → horizontal distance
- Perpendicular → height
- Hypotenuse → line of sight
🔢 Formula
Formula Selection Strategy
\[ \begin{aligned}\text{Choose ratio based}& \text{ on known and required quantities:}\\\\
\tan \theta &= \frac{\text{Height}}{\text{Base}} \quad (\text{Most used})\\\\
\sin \theta &= \frac{\text{Height}}{\text{Hypotenuse}}\\\\
\cos \theta &= \frac{\text{Base}}{\text{Hypotenuse}}
\end{aligned} \]
🗺️ Roadmap
Step-by-Step Problem Solving Roadmap
- Read the question carefully and identify given values
- Draw a labeled right triangle
- Mark angle of elevation/depression clearly
- Identify known and unknown sides
- Select appropriate trigonometric ratio
- Substitute values and solve
- Write final answer with proper unit
✏️ Example
Illustrative Examples (Board Level + Conceptual)
A pole casts a shadow of 10 m when the angle of elevation of the sun is \(45^\circ\). Find the height of the pole.
💡 Concept
Shadow = base, height = perpendicular
\[ \begin{aligned} \tan 45^\circ &= \frac{h}{10}\\ 1 &= \frac{h}{10} \\\Rightarrow h &= 10 \text{ m} \end{aligned} \]
The angle of elevation of a tower from a point is \(30^\circ\). If the height is 20 m, find the distance of the point from the base.
\[ \begin{aligned} \tan 30^\circ &= \frac{20}{d}\\ \frac{1}{\sqrt{3}} &= \frac{20}{d}\\ \Rightarrow d &= 20\sqrt{3} \end{aligned} \]
📐 Derivation
Underlying Derivation Logic
Every single triangle problem reduces to:
\[ \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} \]
By rearranging:
- Height = Base × tan θ
- Base = Height ÷ tan θ
⚡ Exam Tip
Exam Tips (High Scoring Strategy)
⚠️ Warning
Common Mistakes to Avoid
📋 Case Study
CBSE Case Study (HOTS Level)
A tree breaks due to storm and its top touches the ground at a point 8 m from its base. If the angle made with the ground is \(30^\circ\), find the original height of the tree.
Approach:
- Form right triangle with broken part
- Use tan relation for broken segment
- Add vertical stump height
This integrates: geometry + trigonometry + logical modeling
🌟 Importance
Importance for CBSE Board Exams
✏️ Example
Example 1 - Height of Tower using Angle of Elevation
🎨 SVG Diagram
Figure
Right Triangle Representation of Tower Problem
❓ Question
Question
A tower stands vertically on the ground. From a point on the ground,
which is 15 m away from the foot of the tower, the angle of elevation
of the top of the tower is \(60^\circ\). Find the height of the tower.
💡 Concept
Concept Used
🗺️ Roadmap
Raodmap
- Draw triangle ABC (right-angled at B)
- Identify:
- BC = base = 15 m
- AB = height (unknown)
- Angle at C = \(60^\circ\)
- Apply \(\tan \theta\)
- Substitute standard value
🧩 Solution
Step by step Solution
- Using trigonometric ratio:
- \[ \tan 60^\circ = \frac{AB}{BC} \]
- Substitute known values:
- \[ \sqrt{3} = \frac{AB}{15} \]
- \[ AB = 15 \times \sqrt{3} \]
- \[ \begin{aligned} AB &= 15 \times 1.732 \\&= 25.98 \\&\approx 26 \text{ m} \end{aligned} \]
✅ Answer
Final Answer
Final Answer: Height of tower ≈ 26 m
📌 Note
Key Insight
⚡ Exam Tip
Exam Tips
⚠️ Warning
Common Mistakes
📋 Case Study
HOTS Variation
If distance is unknown and height is given, reverse the formula:
\[ \text{Distance} = \frac{\text{Height}}{\tan \theta} \]
✏️ Example
Example 2 – Ladder Problem (Height & Distance Application)
🎨 SVG Diagram
Figure
Ladder Inclined at 60° to Reach Repair Point
❓ Question
Question
An electrician needs to repair a point on a pole of height 5 m, located 1.3 m below the top.
A ladder is inclined at \(60^\circ\) to the horizontal. Find:
1. Length of the ladder
2. Distance of the foot of the ladder from the pole
1. Length of the ladder
2. Distance of the foot of the ladder from the pole
💡 Concept
Concept Used
🗺️ Roadmap
Problem Solving Roadmap
- Convert real scenario into triangle
- Find effective height
- Use tan for base
- Use sin for ladder length
🧩 Solution
Step by step solution
- Initial Calculation
Required height:
\[ AB = 5 - 1.3 = 3.7 \text{ m} \]
- Distance from Pole (Base)
- \[ \tan 60^\circ = \frac{AB}{BC} \]
- \[ \sqrt{3} = \frac{3.7}{BC} \]
- \[ \begin{aligned} BC &= \frac{3.7}{\sqrt{3}} \\&= \frac{3.7}{1.73} \\&\approx 2.14 \text{ m} \end{aligned} \]
- Distance of ladder from pole ≈ 2.14 m
- Length of Ladder (Hypotenuse)
- \[ \sin 60^\circ = \frac{AB}{AC} \]
- \[ \frac{\sqrt{3}}{2} = \frac{3.7}{AC} \]
- \[ \begin{aligned} AC &= \frac{3.7 \times 2}{\sqrt{3}} \\&= \frac{7.4}{1.73} \\&\approx 4.28 \text{ m} \end{aligned} \]
✅ Answer
Final Answer
Length of ladder ≈ 4.28 m
📌 Note
Why Two Ratios Are Used?
⚡ Exam Tip
Exam Tips
⚠️ Warning
Common Mistakes
📋 Case Study
HOTS Extension
If angle changes, how does ladder length vary?
→ Smaller angle → longer ladder required
✏️ Example
Example 3 – Height of Chimney with Observer’s Eye Level
🎨 SVG Diagram
Figure
Observer’s Eye Level Adjustment in Trigonometry
❓ Question
Question
An observer 1.5 m tall is 28.5 m away from a chimney. The angle of elevation
of the top of the chimney from her eyes is \(45^\circ\). Find the height of the chimney.
💡 Concept
Concept Used
🗺️ Roadmap
Solution Roadmap
- Draw triangle from observer’s eye
- Consider only vertical difference (not full height)
- Apply tan θ
- Add observer’s height at the end
🧩 Solution
Step by step Solution
- Let AB be the height of chimney above eye level.
- Using trigonometric ratio:
- \[\tan 45^\circ = \frac{AB}{BC}\]
- \[ \begin{aligned} 1 &= \frac{AB}{28.5}\\ \Rightarrow AB &= 28.5 \text{ m} \end{aligned} \]
- Total height of chimney:
- \[\text{Height} = AB + \text{observer height}\]
- \[= 28.5 + 1.5 = 30 \text{ m}\]
✅ Answer
Final Answer
Final Answer: Height of chimney = 30 m
📌 Note
Why Add Observer Height?
⚡ Exam Tip
Exam Tips
⚠️ Warning
Common Mistakes
📋 Case Study
HOTS Insight
If observer height increases, calculated triangle height decreases but total height remains constant.
✏️ Example
Example 4 – Two Angles of Elevation (Building + Flagstaff)
🎨 SVG Diagram
Figure
Two Triangles Formed by Different Angles of Elevation
❓ Question
Question
From a point P on the ground, the angle of elevation of the top of a 10 m building is \(30^\circ\).
A flagstaff is mounted on top of the building and the angle of elevation of its top is \(45^\circ\).
Find:
1. Length of the flagstaff
2. Distance of the building from point P
1. Length of the flagstaff
2. Distance of the building from point P
💡 Concept
Concept Used
🗺️ Roadmap
Solution Roadmap
- Use smaller angle (30°) → find base AP
- Use larger angle (45°) → find total height
- Subtract building height → flagstaff length
🧩 Solution
Step by step solution
- Distance of Building from P
- \[\tan 30^\circ = \frac{10}{AP}\]
- \[\frac{1}{\sqrt{3}} = \frac{10}{AP}\]
- \[ \begin{aligned} AP &= 10\sqrt{3} \\&= 10 \times 1.732 \\&= 17.32 \text{ m} \end{aligned} \]
- Distance of building from P = 17.32 m
- Total Height (Building + Flagstaff)
- \[\tan 45^\circ = \frac{AD}{AP}\]
- \[ \begin{aligned} 1 &= \frac{AD}{17.32}\\ \Rightarrow AD &= 17.32 \text{ m} \end{aligned} \]
- Length of Flagstaff
- \[\text{Flagstaff} = AD - AB\]
- \[ = 17.32 - 10 = 7.32 \text{ m} \]
✅ Answer
Final Answer
Length of flagstaff = 7.32 m
📌 Note
Why This Method Works
⚡ Exam Tip
Exam Tips
⚠️ Warning
Common Mistakes
📋 Case Study
HOTS Insight
Larger angle → steeper line of sight → higher point observed. This principle is key in comparing multiple elevations.
📄 Example
Example 5 – Shadow Length Variation with Sun’s Altitude
🎨 SVG Diagram
Figure
Shadow Length Increases as Sun’s Altitude Decreases
❓ Question
Question
The shadow of a tower is 40 m longer when the Sun’s altitude is \(30^\circ\) than when it is \(60^\circ\).
Find the height of the tower.
💡 Concept
Concept Used
🗺️ Roadmap
Assumption & Variable Setup
- Let height of tower = \(h\)
- Shorter shadow (at \(60^\circ\)) = \(x\)
- Longer shadow (at \(30^\circ\)) = \(x + 40\)
🧩 Solution
Step by step solution
- Forming Equations
- Sun altitude = \(60^\circ\)
- \[ \begin{align} \tan 60^\circ &= \frac{h}{x}\\ \Rightarrow h &= x\sqrt{3} \tag{1} \end{align} \]
- Sun altitude = \(30^\circ\)
- \[\tan 30^\circ = \frac{h}{x + 40}\]
- \[\frac{1}{\sqrt{3}} = \frac{h}{x + 40}\tag{2}\]
- Solving the Equations
- Substitute (1) into second equation:
- \[\frac{1}{\sqrt{3}} = \frac{x\sqrt{3}}{x + 40}\]
- \[x + 40 = 3x\]
- \[ 2x = 40 \Rightarrow x = 20\]
- Final Height Calculation
- \[h = x\sqrt{3} = 20\sqrt{3}\]
- \[h = 20 \times 1.732 = 34.64 \text{ m}\]
✅ Answer
Final Answer
Final Answer: Height of tower = 34.64 m
💡 Concept
Conceptual Insight
⚡ Exam Tip
Exam Tips
⚠️ Warning
Common Mistakes
📋 Case Study
HOTS Insight
These problems test your ability to compare two situations and form equations — a key algebra + trigonometry integration skill.
✏️ Example
Example 6 – Angles of Depression (Two Buildings Problem)
🎨 SVG Diagram
Figure
Angles of Depression Equal Corresponding Elevation Angles
❓ Question
Question
The angles of depression of the top and bottom of an 8 m building from the top of another building
are \(30^\circ\) and \(45^\circ\), respectively. Find:
1. Height of the taller building
2. Distance between the buildings
1. Height of the taller building
2. Distance between the buildings
💡 Concept
Concept Used
🗺️ Roadmap
Assumption
- Let distance between buildings = \(d\)
- Let extra height above 8 m building = \(x\)
- Total height of taller building = \(x + 8\)
🧩 Solution
Step by step solution
- Forming Equations
- From bottom (45°)
- \[ \begin{align} \tan 45^\circ &= \frac{x + 8}{d}\\ \Rightarrow 1 &= \frac{x + 8}{d}\\ \Rightarrow d &= x + 8 \tag{1} \end{align} \]
- From top (30°)
- \[ \begin{align} \tan 30^\circ = \frac{x}{d}\\ \Rightarrow \frac{1}{\sqrt{3}} = \frac{x}{d}\\ \Rightarrow d = x\sqrt{3} \tag{2} \end{align} \]
- Solving Equations
Equating (1) and (2): - \[x + 8 = x\sqrt{3}\]
- \[x(\sqrt{3} - 1) = 8\]
- \[x = \frac{8}{\sqrt{3} - 1}\]
- Rationalizing
- \[ \begin{aligned} x &= \frac{8(\sqrt{3} + 1)}{2} \\&= 4(\sqrt{3} + 1) \end{aligned} \]
✅ Answer
Final Answer
- Height of taller building
- \[ \begin{aligned} x + 8 &= 4(\sqrt{3} + 1) + 8 \\&= 12 + 4\sqrt{3} \\&= 4(3 + \sqrt{3}) \text{ m} \end{aligned} \]
- Distance between buildings
- \[ \begin{aligned} d &= x + 8 \\&= 4(3 + \sqrt{3}) \text{ m} \end{aligned} \]
- Height = Distance = \(4(3 + \sqrt{3})\) m
📌 Note
Key Insight
⚡ Exam Tip
Exam Tips
⚠️ Warning
Common Mistakes
📋 Case Study
HOTS Insight
Multi-angle problems test your ability to form simultaneous equations — a key bridge between algebra and trigonometry.
📌 Note
Advanced Concepts & Theory – Applications of Trigonometry
NCERT · Class X · Mathematics · Chapter 9
Some Applications of Trigonometry
Heights & Distances — Angles of Elevation & Depression — Complete AI Learning Engine
Core Concepts — Chapter 9
What This Chapter is About
Chapter 9 applies the trigonometric ratios from Chapters 8 to real-world problems involving heights and distances. The core idea: in a right triangle formed by an observer, an object, and the ground, we know one side (horizontal distance or height) and one angle, and we use tan, sin, or cos to find the unknown side.
📐
Angle of Elevation
When you look UP at an object from horizontal ground
Definition
The angle of elevation is the angle formed between the horizontal line through the observer's eye and the line of sight to an object above. It is always measured upward from the horizontal.
tan θ = h / d → h = d · tan θ | sin θ = h / hypotenuse | cos θ = d / hypotenuse
📉
Angle of Depression
When you look DOWN at an object from an elevated position
Definition
The angle of depression is the angle formed between the horizontal line through the observer's eye and the line of sight to an object below. It is always measured downward from the horizontal. By alternate interior angles, the angle of depression from A to B equals the angle of elevation from B to A.
tan θ = h / d (same formula!) | Angle of depression from top = Angle of elevation from bottom (alternate angles)
📌 Key Principle: Both angle of elevation and angle of depression problems reduce to the same right triangle. The formula is always
tan(θ) = opposite / adjacent = height / horizontal distance.
The only difference is who is observing and from where.
Problem-Solving Framework — 4 Steps
- Step 1 — Draw: Always draw a neat diagram. Mark the right angle, the given angle (elevation/depression), and label known/unknown sides.
- Step 2 — Identify: Determine which two sides are involved relative to the given angle (opposite, adjacent, hypotenuse).
- Step 3 — Apply: Choose the correct ratio — usually tan (when height and horizontal distance are involved), sin or cos (when hypotenuse/line of sight is involved).
- Step 4 — Substitute: Plug in the standard value (tan 30° = 1/√3, tan 45° = 1, tan 60° = √3), solve the equation, and rationalise if needed.
Complete Formula Reference — Chapter 9
🔺
Core Trigonometric Ratios for Right Triangles
The three ratios used in every heights-and-distances problem
tan θ = Opposite / Adjacent
Tangent — Most Used
= height / horizontal distance | Used in ~85% of Chapter 9 problems
sin θ = Opposite / Hypotenuse
Sine — Line of Sight problems
= height / line of sight | Used when slant distance is given
cos θ = Adjacent / Hypotenuse
Cosine — Horizontal reach
= horizontal / line of sight | Used when slant distance is given
📊
Derived Height & Distance Formulas
Direct-use formulas for all standard configurations
h = d · tan θ
Height from Angle of Elevation
h = height of object, d = horizontal distance, θ = angle of elevation
d = h / tan θ = h · cot θ
Distance from Angle of Depression
h = observer height, θ = angle of depression
h = d(tan α − tan β) / 1
Two-Angle Problem (same side)
α = angle near, β = angle far, d = distance between two points of observation
h = d · tan α · tan β / (tan α − tan β)
Two-Angle Problem (moving observer)
Observer moves d metres closer; angles change from β to α
Shadow Length = h / tan θ
Shadow Problems
θ = angle of elevation of sun, h = height of object
L = h / sin θ
Length of Line of Sight (rope/ladder)
L = slant length, h = vertical height, θ = angle with horizontal
📋
Standard Trigonometric Values — Quick Reference
Memorise these — every Chapter 9 problem uses exactly one of these angles
| Angle θ | sin θ | cos θ | tan θ | h = d·tan θ (d=10) |
|---|---|---|---|---|
| 0° | 0 | 1 | 0 | 0 m |
| 30° | 1/2 | √3/2 | 1/√3 ≈ 0.577 | 10/√3 ≈ 5.77 m |
| 45° | 1/√2 | 1/√2 | 1 | 10 m |
| 60° | √3/2 | 1/2 | √3 ≈ 1.732 | 10√3 ≈ 17.32 m |
| 90° | 1 | 0 | undefined | ∞ |
✦ In NCERT Chapter 9, only 30°, 45°, and 60° appear as angles of elevation/depression. 90° and 0° are theoretical limits.
🧮
Two-Tower & Two-Observer Configurations
Advanced patterns from the harder Chapter 9 problems
| Configuration | Setup | Key Formula |
|---|---|---|
| Two angles, one observer | Observer sees top of tower at α, bottom at β (from far) | h·(cot β − cot α) = horizontal spacing |
| Observer moves closer | Angle α at distance d₁, angle β at distance d₂ (d₂ < d₁) | h = (d₁−d₂)·tan α·tan β / (tan β − tan α) |
| Two towers, between them | Angles of elevation α and β from a point between the towers | h₁/tan α + h₂/tan β = total distance |
| Flagpole on building | Angles to bottom and top of flagpole from ground | Pole height = d(tan β − tan α) |
Tips, Tricks & Smart Strategies
✨
Expert Strategies for Chapter 9 Problems
Exam-tested techniques for heights and distances
1
Draw First — Always
Before writing a single equation, draw the diagram. Mark the observer (eye-level or at a point), the object, the right angle (at the base), the given angle, and label all known sides. A clear diagram eliminates 90% of errors in this chapter.
2
tan is Your Best Friend
tan 45° = 1 → height = d → clean integer answer
tan 60° = √3 → height = d√3 → very common answer form
For all problems where you know the horizontal distance (or need it) and the height: use tan θ = height / horizontal distance. You almost never need sin or cos unless the slant distance (line of sight / rope / ladder length) is given or asked.
tan 30° = 1/√3 → height = d/√3 → rationalise: d√3/3tan 45° = 1 → height = d → clean integer answer
tan 60° = √3 → height = d√3 → very common answer form
3
Alternate Angles Theorem — Depression = Elevation
When an observer at height h looks down at an angle of depression θ to a point on the ground, the horizontal line and ground are parallel. By alternate interior angles, the angle of elevation from the ground point to the observer also equals θ. This lets you draw the right triangle with the right angle at the base — not at the observer's height.
4
Rationalisation — Never Leave √ in Denominator
d = 20√3 m ← leave in surd form (exact answer preferred)
NCERT always expects rationalised answers. When you get h = d/√3, multiply numerator and denominator by √3 to get h = d√3/3.
h = 50/√3 = 50√3/3 ≈ 28.87 m ← correct formd = 20√3 m ← leave in surd form (exact answer preferred)
5
Two-Point Problems — Use Simultaneous Equations
From B: tan β = h/(x+d) → h = (x+d)·tan β …(ii)
Equate (i) and (ii), solve for x, then find h.
When angles are given from two different positions (e.g., A and B are 100 m apart, looking at tower top), set up two equations with two unknowns (height h and one distance). Divide or subtract to eliminate one unknown.
From A: tan α = h/x → h = x·tan α …(i)From B: tan β = h/(x+d) → h = (x+d)·tan β …(ii)
Equate (i) and (ii), solve for x, then find h.
6
Shadow Problems — Sun Angle is Elevation Angle
→ shadow length = height / tan(sun angle)
In shadow problems, the angle of elevation of the sun is the angle from the tip of the shadow to the top of the object. The right triangle is: vertical side = object height, horizontal side = shadow length, angle at shadow tip = sun's elevation angle.
tan(sun angle) = height of object / shadow length→ shadow length = height / tan(sun angle)
7
Flagpole-on-Building Problems — Two Triangles Share Base
where α = angle to building top, β = angle to pole top
When a flagpole stands on top of a building: draw two separate right triangles from the observer. Both triangles share the same horizontal base (distance from observer to building). Set up two tan equations — one for the building angle, one for the top-of-pole angle — and subtract.
Pole height = d·(tan β − tan α)where α = angle to building top, β = angle to pole top
8
Point Between Two Objects — Work from Both Sides
If a person stands between two towers and angles to tops of both towers are given: set the person's distance from one tower as x and from the other as (d − x). Write two height equations and equate if both towers are the same height, or solve the system if heights differ.
9
Standard Answer Forms to Memorise
If angle is 45°: height = distance (clean) — e.g., h = 50 m
If angle is 60°: height = distance × √3 — e.g., h = 50√3 m
If angle is 30°: height = distance / √3 = distance·√3/3
Most Chapter 9 answers come in these forms — recognise them instantly:
Common results: 10√3, 20√3, 50√3/3, 100/√3 = 100√3/3If angle is 45°: height = distance (clean) — e.g., h = 50 m
If angle is 60°: height = distance × √3 — e.g., h = 50√3 m
If angle is 30°: height = distance / √3 = distance·√3/3
Common Mistakes & Misconceptions
⚠️
Critical Errors in Heights & Distances
Each error is based on real student mistakes in board exams
Using sin or cos when tan is the right choice
✗ Wrong: Height known, horizontal dist. needed → uses sin θ = h / LS (involves unknown hypotenuse)
✓ Right: When height and horizontal distance are the two sides, always use tan θ = h / d
sin and cos involve the hypotenuse (slant/line-of-sight). Use them only when line-of-sight length is given or required. tan is used when vertical height and horizontal distance are the known/unknown pair.
Measuring angle from the wrong reference line
✗ Wrong: Measuring the angle of depression from the vertical instead of the horizontal
✓ Right: Angles of elevation AND depression are always measured from the HORIZONTAL line
Both elevation and depression angles are measured from the horizontal at the observer's eye level. Never from the vertical. A common error is drawing the angle at the vertical side of the tower.
Not using the alternate angles property for depression
✗ Wrong: In depression problems, taking the angle at the top of the cliff instead of forming a right triangle from the base
✓ Right: Use alternate angles — angle of depression from top = angle of elevation from ground object. Form the right triangle with right angle at the base of the cliff.
Students often form a right triangle at the top of the cliff (making the horizontal side = h, which is wrong). The correct right triangle has the vertical leg = cliff height and horizontal leg = ground distance.
Forgetting observer's eye height
✗ Wrong: "A man 1.8 m tall observes a tower at 45°. Distance = 50 m. Height = 50·tan45° = 50 m" (ignoring eye height)
✓ Right: Height of tower = 50·tan45° + 1.8 = 50 + 1.8 = 51.8 m (add observer's eye height)
If the problem states the observer's height, the line of sight starts from the eye level, not ground level. The calculated height must be added to the observer's eye height to get the tower's total height from the ground.
Leaving answers with √ in the denominator
✗ Wrong: Height = 100/√3 m (not rationalised)
✓ Right: Height = 100/√3 × √3/√3 = 100√3/3 m
NCERT solutions always rationalise surds. 100/√3 and 100√3/3 are mathematically equal, but the second form is the expected answer format.
Confusing which angle goes with which observer position
✗ Wrong: In two-point problems, using the larger angle with the farther distance
✓ Right: Larger angle of elevation → closer observer. Smaller angle → farther observer.
As you move closer to a tower, the angle of elevation increases. So if observer moves from B to A (closer) and angle increases from β to α: α > β and d(B) > d(A). Always verify this makes geometric sense.
Setting up the wrong right angle in the triangle
✗ Wrong: Placing the right angle at the observer's position instead of the base of the object
✓ Right: The right angle is always at the foot of the perpendicular — the base of the tower/pole, where the vertical meets the horizontal ground
The right triangle in elevation/depression problems has: one vertex at the observer, one at the foot of the object, one at the top of the object. The 90° angle is always at the foot of the object (vertical meets horizontal ground).
Arithmetic errors with √3
✗ Wrong: 20√3 ÷ √3 = 20√3 · √3 = 20·3 = 60 (but written as h = 60/√3 = 20√3 then stops here)
✓ Right: If h·√3 = 60, then h = 60/√3 = 60√3/3 = 20√3 m (verify by checking 20√3·√3 = 20·3 = 60 ✓)
When solving equations with √3, always write out every step: 60/√3 → rationalize → 60√3/3 → simplify → 20√3. Skipping steps leads to wrong numerical answers.
Concept-Building Practice Questions
Q 01
A tower stands on a horizontal plane. From a point 80 m away from its foot, the angle of elevation of the top is 60°. Find the height of the tower.
Elevation
▼
DIAGRAM
Let the tower be AB (A = top, B = foot). Observer at C, angle ACB = 60°, BC = 80 m, right angle at B.
FORMULA
tan(∠ACB) = AB/BC → tan 60° = h/80
SUBSTITUTE
√3 = h/80 → h = 80√3 m
VERIFY
80√3 ≈ 80 × 1.732 = 138.56 m. Reasonable for the 60° elevation. ✓
Height of tower = 80√3 m ≈ 138.56 m
Q 02
A ladder 15 m long makes an angle of 60° with the wall. Find how high up the wall the ladder reaches.
ElevationFind Height
▼
SETUP
Ladder = hypotenuse = 15 m. Angle with the wall = 60°, so angle with the ground = 30°. Height reached = side adjacent to ground angle = side opposite to wall angle.
CHOOSE RATIO
cos(angle with wall) = height/hypotenuse → cos 60° = h/15
OR: sin(angle with ground) = h/15 → sin 30° = h/15
OR: sin(angle with ground) = h/15 → sin 30° = h/15
CALCULATE
cos 60° = 1/2 → h = 15 × 1/2 = 7.5 m
VERIFY
Ground distance = 15·sin60° = 15·√3/2 = 7.5√3 m. Check: h² + d² = 7.5² + (7.5√3)² = 56.25 + 168.75 = 225 = 15² ✓
Height reached on wall = 7.5 m
Q 03
From the top of a cliff 120 m high, the angle of depression of a boat in the sea is 30°. How far is the boat from the foot of the cliff?
Depression
▼
ALTERNATE ANGLES
Angle of depression from cliff top = 30°. By alternate angles (horizontal lines are parallel), angle of elevation from boat to cliff top = 30°.
RIGHT TRIANGLE
Right triangle: vertical side = 120 m (cliff height), horizontal side = d (boat distance), angle at boat = 30°.
FORMULA
tan 30° = 120/d → 1/√3 = 120/d → d = 120√3 m
ANSWER
d = 120√3 ≈ 120 × 1.732 = 207.84 m
Distance of boat from foot of cliff = 120√3 m ≈ 207.84 m
Q 04
The angles of elevation of the top of a tower from two points P and Q, 9 m apart on the same side, are 30° and 60° respectively. Prove that the height of the tower is 3√3 m.
Two Angles
▼
SETUP
Let tower height = h, distance from nearer point Q to tower = x m. Then P is farther, distance = (x + 9) m. Angle at P = 30°, angle at Q = 60°.
EQUATION 1
From Q (nearer, larger angle): tan 60° = h/x → √3 = h/x → h = √3 x …(i)
EQUATION 2
From P (farther, smaller angle): tan 30° = h/(x+9) → 1/√3 = h/(x+9) → h = (x+9)/√3 …(ii)
EQUATE
√3 x = (x+9)/√3 → 3x = x + 9 → 2x = 9 → x = 9/2 = 4.5 m
FIND h
h = √3 × 4.5 = 4.5√3 = 9√3/2 m. Wait — let me re-examine. Q is 9 m apart from P, P is farther.
Actually: if P gives 30° and Q gives 60°, and PQ = 9 m, and Q is nearer (60° > 30°):
Let QB = x. PB = x + 9. From Q: h = x√3. From P: h = (x+9)/√3.
→ x√3 = (x+9)/√3 → 3x = x+9 → x = 4.5 → h = 4.5√3 = 9√3/2 m.
For h = 3√3: need PQ differently. If PB = x, QB = x−9: tan60 = h/(x−9), tan30 = h/x → x = 9√3/2... Let's use: h² = 9·tan30·tan60 = 9·(1/√3)·√3 = 9 → h = 3 m? No.
Geometric mean: h² = d₁·d₂ where d₁,d₂ are distances → h = √(d₁·d₂). With d₁·d₂ = x(x+9) and h=x√3=√3x → (√3x)² = x(x+9)... Actually for h=3√3: using h=3√3, from Q: x=h/√3=3, from P: x+9=h√3=9. So PQ = 9−3 = 6 m, not 9. The problem might intend PQ=9m with the answer being 9√3/2. Let's present the correct derivation.
Actually: if P gives 30° and Q gives 60°, and PQ = 9 m, and Q is nearer (60° > 30°):
Let QB = x. PB = x + 9. From Q: h = x√3. From P: h = (x+9)/√3.
→ x√3 = (x+9)/√3 → 3x = x+9 → x = 4.5 → h = 4.5√3 = 9√3/2 m.
For h = 3√3: need PQ differently. If PB = x, QB = x−9: tan60 = h/(x−9), tan30 = h/x → x = 9√3/2... Let's use: h² = 9·tan30·tan60 = 9·(1/√3)·√3 = 9 → h = 3 m? No.
Geometric mean: h² = d₁·d₂ where d₁,d₂ are distances → h = √(d₁·d₂). With d₁·d₂ = x(x+9) and h=x√3=√3x → (√3x)² = x(x+9)... Actually for h=3√3: using h=3√3, from Q: x=h/√3=3, from P: x+9=h√3=9. So PQ = 9−3 = 6 m, not 9. The problem might intend PQ=9m with the answer being 9√3/2. Let's present the correct derivation.
CORRECT ANSWER
With PQ = 9 m: h = 9√3/2 m. For h = 3√3: PQ must be 6 m, distance from nearer = 3 m, from farther = 9 m. Verify: tan60° = 3√3/3 = √3 ✓, tan30° = 3√3/9 = √3/3 = 1/√3 ✓
Height = 9√3/2 m (for PQ=9m) | Concept: h = √(d₁·d₂·tan α·tan β)
Q 05
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
Shadow
▼
KEY PRINCIPLE
At the same time of day, the sun's angle of elevation is the same for all objects. So the ratio of height to shadow length is constant.
PROPORTION
Height of pole / Shadow of pole = Height of tower / Shadow of tower
6/4 = H/28
6/4 = H/28
SOLVE
H = 28 × 6/4 = 28 × 3/2 = 42 m
Height of tower = 42 m
Q 06
From a point on the ground, the angles of elevation of the bottom and top of a flagpole fixed atop a building 20 m high are 45° and 60° respectively. Find the length of the flagpole.
Two AnglesCombined
▼
SETUP
Building height = AB = 20 m, flagpole = BC on top. Observer at O on ground, OA = d (horizontal). ∠AOB = 45°, ∠AOC = 60°.
FIND d
tan 45° = AB/OA = 20/d → 1 = 20/d → d = 20 m
FIND OC
tan 60° = OC/OA = OC/20 → √3 = OC/20 → OC = 20√3 m
(OC = total height = building + flagpole)
(OC = total height = building + flagpole)
FLAGPOLE LENGTH
Flagpole BC = OC − OB = 20√3 − 20 = 20(√3 − 1) m
NUMERICAL
20(√3−1) ≈ 20(1.732−1) = 20 × 0.732 = 14.64 m
Length of flagpole = 20(√3 − 1) m ≈ 14.64 m
Q 07
Two boats are on opposite sides of a lighthouse of height 100 m. The angles of depression of the two boats from the top are 45° and 30°. Find the distance between the two boats.
DepressionCombined
▼
SETUP
Lighthouse AB = 100 m. Boat C on one side, depression angle 45°. Boat D on other side, depression angle 30°.
DISTANCE TO C
tan 45° = AB/BC = 100/BC → 1 = 100/BC → BC = 100 m
DISTANCE TO D
tan 30° = AB/BD = 100/BD → 1/√3 = 100/BD → BD = 100√3 m
TOTAL DISTANCE
CD = BC + BD = 100 + 100√3 = 100(1 + √3) m
NUMERICAL
100(1 + √3) ≈ 100(1 + 1.732) = 100 × 2.732 = 273.2 m
Distance between boats = 100(1 + √3) m ≈ 273.2 m
Q 08
A man observes a tower at 30° elevation. He walks 20 m towards the tower and now the elevation becomes 60°. Find the height of the tower and original distance from the tower.
Two Angles
▼
SETUP
Tower AB = h. Original observer position C: angle = 30°, BC = d. After walking 20 m: position D: angle = 60°, BD = d − 20.
EQUATION 1
From C: tan 30° = h/d → 1/√3 = h/d → d = h√3 …(i)
EQUATION 2
From D: tan 60° = h/(d−20) → √3 = h/(d−20) → d−20 = h/√3 …(ii)
SUBSTITUTE
From (i): d = h√3. Substitute in (ii): h√3 − 20 = h/√3
h√3 − h/√3 = 20 → h(√3 − 1/√3) = 20 → h·(3−1)/√3 = 20 → h·2/√3 = 20 → h = 10√3 m
h√3 − h/√3 = 20 → h(√3 − 1/√3) = 20 → h·(3−1)/√3 = 20 → h·2/√3 = 20 → h = 10√3 m
FIND d
d = h√3 = 10√3 × √3 = 10 × 3 = 30 m
VERIFY
From C: tan θ = 10√3/30 = √3/3 = 1/√3 = tan 30° ✓
From D: tan θ = 10√3/10 = √3 = tan 60° ✓
From D: tan θ = 10√3/10 = √3 = tan 60° ✓
Height = 10√3 m ≈ 17.32 m | Original distance = 30 m
Q 09
A window in a building is 8 m above the ground. From the window, the angle of elevation of a nearby building's top is 60° and the angle of depression of its foot is 45°. Find the height of the nearby building and its distance from the window-building.
DepressionCombined
▼
SETUP
Window W is 8 m above ground. Nearby building top T, foot F. WF and WT are lines of sight. Let horizontal distance between buildings = d. Let TW's height above W = x (upward part).
ANGLE OF DEPRESSION
Depression to foot F = 45°. The right triangle: horizontal d, vertical = 8 m (window height).
tan 45° = 8/d → 1 = 8/d → d = 8 m
tan 45° = 8/d → 1 = 8/d → d = 8 m
ANGLE OF ELEVATION
Elevation to top T = 60°. The right triangle: horizontal d = 8 m, vertical rise above window = x.
tan 60° = x/8 → √3 = x/8 → x = 8√3 m
tan 60° = x/8 → √3 = x/8 → x = 8√3 m
TOTAL HEIGHT
Height of nearby building = window height + x = 8 + 8√3 = 8(1 + √3) m
NUMERICAL
8(1+√3) ≈ 8 × 2.732 = 21.86 m
Distance = 8 m | Height of nearby building = 8(1+√3) m ≈ 21.86 m
Q 10
A straight highway leads to the foot of a tower. A man standing at the top sees a car at angle of depression 30°. The car approaches the tower and after 2 minutes the angle of depression is 60°. How long will it take to reach the foot of the tower?
DepressionTwo Angles
▼
SETUP
Tower height = h. At first: car at C, depression 30°, distance BC = d₁. After 2 min: car at D, depression 60°, distance BD = d₂.
DISTANCES
tan 30° = h/d₁ → d₁ = h√3
tan 60° = h/d₂ → d₂ = h/√3 = h√3/3
tan 60° = h/d₂ → d₂ = h/√3 = h√3/3
DISTANCE COVERED
CD = d₁ − d₂ = h√3 − h√3/3 = h√3(1 − 1/3) = 2h√3/3
REMAINING
DB = d₂ = h√3/3 = h/√3
CD = 2h√3/3 | DB = h√3/3
Ratio DB/CD = (h√3/3)/(2h√3/3) = 1/2
CD = 2h√3/3 | DB = h√3/3
Ratio DB/CD = (h√3/3)/(2h√3/3) = 1/2
TIME
CD was covered in 2 minutes. DB = CD/2 → time = 2/2 = 1 minute
Time to reach the foot of the tower = 1 minute
Q 11
At what angle does the sun's elevation need to be so that a pole of height h casts a shadow of length h/√3? Also find the shadow when the sun is at 45°.
ShadowFind Angle
▼
PART 1
tan θ = height / shadow = h / (h/√3) = h × √3/h = √3
tan θ = √3 → θ = 60°
tan θ = √3 → θ = 60°
PART 2
At 45°: tan 45° = h / shadow → 1 = h / shadow → shadow = h
INSIGHT
Shadow length = h/tan θ. As sun rises (θ increases), shadow gets shorter. At 90° (sun directly overhead): shadow = 0. At 0°: shadow is infinite.
Sun elevation = 60° for shadow = h/√3 | At 45°, shadow = h (same as pole height)
Q 12
Two poles of equal height stand on either side of a road 80 m wide. At a point P between the poles, the elevations to the tops are 60° and 30°. Find the height of the poles and the position of point P.
CombinedTwo Angles
▼
SETUP
Poles AB = CD = h. Road width AC = 80 m. P is on AC, AP = x, PC = 80−x. Angle of elevation to B (top of AB) = 60°, to D (top of CD) = 30°.
EQUATION 1
tan 60° = h/AP = h/x → √3 = h/x → h = x√3 …(i)
EQUATION 2
tan 30° = h/PC = h/(80−x) → 1/√3 = h/(80−x) → h = (80−x)/√3 …(ii)
EQUATE
x√3 = (80−x)/√3 → 3x = 80−x → 4x = 80 → x = 20 m
FIND h
h = 20√3 m
VERIFY
PC = 80−20 = 60 m. h/PC = 20√3/60 = √3/3 = 1/√3 = tan 30° ✓
Height of poles = 20√3 m ≈ 34.64 m | P is 20 m from first pole, 60 m from second
Step-by-Step AI Solver
Select a problem category and scenario below. The solver draws on the full set of Chapter 9 formulas — angle of elevation, angle of depression, two-angle problems, and shadow problems — and works through each step with the trigonometric identity used at every stage clearly labelled.
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ACADEMIA AETERNUM
तमसो मा ज्योतिर्गमय · Est. 2025
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Class 10 Maths Ch 9 Notes: Applications of Trigonometry (Easy)
Class 10 Maths Ch 9 Notes: Applications of Trigonometry (Easy) — Complete Notes & Solutions · academia-aeternum.com
Trigonometry becomes truly meaningful when it steps beyond theoretical ratios and begins to illuminate how we measure the world around us. Chapter 9, Some Applications of Trigonometry, introduces learners to the practical side of this discipline by showing how heights and distances can be determined without physically reaching an object. Whether estimating the altitude of a hill, the height of a building, the distance of a ship from a lighthouse, or the position of a kite in the sky, students…
🎓 Class 10
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