Class 10 Maths Chapter 9: Applications of Trigonometry Notes & Formulas

Exam Weightage & Blueprint

Total: ~4-6 Marks

This chapter falls under Unit V: Trigonometry (12 marks total). As per the latest syllabus: Simple problems on heights and distances. Problems should not involve more than two right triangles. Angles of elevation / depression should be only 30°, 45°, 60°.

Question Type	Marks	Frequency	Focus Topic
MCQ	1	Low	Simple height finding ($\tan \theta$)
Short Answer	3	Medium	Single Triangle Problems (Broken Tree, etc.)
Long Answer / Case Study	4-5	Very High	Double Triangle Problems (Two ships, Building & Tower)

CBSE 2026–27 Syllabus Status: Some Applications of Trigonometry

Unit V: Trigonometry • 4–5 Marks

This chapter is strictly structured according to the CBSE Rationalised Mathematics Syllabus. Focus your board preparation only on prescribed topics and avoid studying deleted sections:

In 2026–27 Board Syllabus:

Deleted / Not in Board Exam:

⏰ Last 24-Hour Checklist

Angle of Elevation: Looking UP from horizontal.
Angle of Depression: Looking DOWN from horizontal.
Key Relation: Angle of Depression = Angle of Elevation (Alternate Angles).

Values: Memorize $\tan 30^°, \tan 45^°, \tan 60^°$.
Diagram Rule: Always mark the right angle and the reference angle.
Root Values: $\sqrt{3} = 1.732$ (Use only if asked).

🔭 Core Theory & Definitions

Line of Sight: The line drawn from the eye of an observer to the point in the object viewed by the observer.

Angles Explained

Angle of Elevation

The angle formed by the line of sight with the horizontal when the point being viewed is above the horizontal level.

Angle of Depression

The angle formed by the line of sight with the horizontal when the point being viewed is below the horizontal level.

Visualising Angles of Elevation & Depression

Angle of Elevation (Looking UP)

Angle of Depression (Looking DOWN)

Important: In Angle of Depression problems, always draw a horizontal line from the observer's eye level first. Then mark the angle downwards.

📐 Essential Formulas & Values

Most questions are solved using the tangent ratio:

$$ \tan \theta = \frac{Perpendicular (Height)}{Base (Distance)} $$

Values to Remember

30-
$\tan 30^^\circ = \frac{1}{\sqrt{3}}$

45-
$\tan 45^^\circ = 1$

60-
$\tan 60^^\circ = \sqrt{3}$

⚡ High-Yield Conceptual Rules (Board MCQs & A-R):

Observer Height Adjustment: If an observer of height $h_0$ views an object of height $H$, the vertical leg in the right triangle is $(H - h_0)$. The total object height is calculated as $\text{height from triangle} + h_0$.
Sun's Altitude vs Shadow Length:
• When angle of elevation $\theta = 45^\circ \implies$ Shadow length = Object height.
• When $\theta = 30^\circ \implies$ Shadow length $= \sqrt{3} \times$ Height (longer shadow).
• As the Sun's altitude $\theta$ increases, shadow length decreases.
✅ As Sun altitude increases ($30^\circ \to 60^\circ$), shadow length decreases
✅ Reflection in water: height above water surface = image depth below surface
Reflection in Lake / Water Surface: The height of a cloud above the water surface is equal to the depth of its reflection below the water surface ($H_{\text{cloud}} = H_{\text{image}}$).

Solved Examples (Board Marking Scheme)

Q1. A tower stands vertically on the ground. From a point 15 m away from the foot, the angle of elevation is $60^^\circ$. Find the height. (2 Marks)

Step 1: Diagram & Given 0.5 Mark

Let AB be the tower ($h$). Point C is 15 m away. $\angle ACB = 60^^\circ$.

Step 2: Apply Ratio 1 Mark

In $\triangle ABC$, $\tan 60^^\circ = \frac{AB}{BC} = \frac{h}{15}$.

Step 3: Solve 0.5 Mark

$\sqrt{3} = \frac{h}{15} \implies h = 15\sqrt{3}$ m.

Q2. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is $60^^\circ$ and the angle of depression of its foot is $45^^\circ$. Determine the height of the tower. (5 Marks)

Step 1: Setup & Diagram 1 Mark

Let AB be the building (7m) and CD be the tower ($H$). Draw horizontal line AE.

$\angle CAE = 60^^\circ$ (Elevation), $\angle EAD = 45^^\circ$ (Depression).

Step 2: Triangle 1 (Lower) 1.5 Marks

In $\triangle ABD$, $\tan 45^^\circ = \frac{AB}{BD} = \frac{7}{BD}$.

$1 = \frac{7}{BD} \implies BD = 7$ m.

Also, $AE = BD = 7$ m.

Step 3: Triangle 2 (Upper) 1.5 Marks

In $\triangle AEC$, $\tan 60^^\circ = \frac{CE}{AE}$.

$\sqrt{3} = \frac{CE}{7} \implies CE = 7\sqrt{3}$ m.

Step 4: Total Height 1 Mark

Height of tower = $CE + ED = 7\sqrt{3} + 7 = 7(\sqrt{3} + 1)$ m.

Previous Year Questions (PYQs)

2023 (Case Study): A kite is flying at a height of 60m. The string makes an angle of $60^^\circ$ with the ground. Find string length.
Ans: Use $\sin 60^° = $\frac{P}{H}$ \implies $\frac{$\sqrt{3}$}{2}$ = $\frac{60}{L}$ \implies L = $\frac{120}{$\sqrt{3}$}$ = 40$\sqrt{3}$$ m.

2020 (3 Marks): 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.
Ans: $h(\cot 30^° - \cot 60^°) = 40 \implies h($\sqrt{3}$ - $\frac{1}{$\sqrt{3}$}$) = 40 \implies h = 20$\sqrt{3}$$ m.

2019 (4 Marks): Two poles of equal heights stand on either side of an 80m wide road. Angles of elevation are $60^^\circ$ and $30^^\circ$. Find height and distances.
Ans: $h = 20\sqrt{3}$ m. Distances are 20m and 60m.

Exam Strategy & Mistake Bank

⚠️ Mistake Bank

Wrong Ratio: Using $\tan$ when hypotenuse (ladder/kite string) is involved. Use $\sin$ or $\cos$ instead.

Angle Place: Marking the angle of depression with the vertical instead of the horizontal.

Calculation: $\sqrt{3} ≈ 1.732$. Use this value only if specified in the question.

💡 Scoring Tips

Diagram is Mandatory: Even if you solve it correctly, no marks for diagram = marks deducted.

Label Vertices: Always label the triangle vertices (A, B, C) and state "Let AB be the tower...".

Double Triangles: In 5-mark questions, you usually get two equations with 'h' and 'x'. Isolate 'x' in one and substitute in the other.

📝 More Solved Board Questions

Q3. A 1.5m tall boy is standing at some distance from a 30m tall building. The angle of elevation from his eyes to the top of the building increases from $30^^\circ$ to $60^^\circ$ as he walks towards the building. Find the distance he walked. 4 Marks

Sol. Height above eye level = $30 - 1.5 = 28.5$ m.

In $\triangle 1$ ($60^^\circ$): $\tan 60^^\circ = \frac{28.5}{x} \implies x = \frac{28.5}{\sqrt{3}}$ m.

In $\triangle 2$ ($30^^\circ$): $\tan 30^^\circ = \frac{28.5}{x+d} \implies \frac{1}{\sqrt{3}} = \frac{28.5}{x+d} \implies x+d = 28.5\sqrt{3}$.

$d = 28.5\sqrt{3} - \frac{28.5}{\sqrt{3}} = \frac{28.5(3-1)}{\sqrt{3}} = \frac{57}{\sqrt{3}} = 19\sqrt{3}$ m.

Answer: $19\sqrt{3}$ m

Q4. As observed from the top of a 100 m high lighthouse from the sea-level, the angles of depression of two ships on opposite sides of the lighthouse are $30^^\circ$ and $45^^\circ$. Find the distance between the two ships. [Use $\sqrt{3} = 1.732$] 5 Marks (CBSE 2023, 2020)

Sol. Let $AB = 100 m$ be the lighthouse. Let ship $C$ be on one side with angle of depression $45^^\circ$ ($\implies \angle ACB = 45^^\circ$), and ship $D$ on the opposite side with angle of depression $30^^\circ$ ($\implies \angle ADB = 30^^\circ$).

In right $\triangle ABC$:

$$\tan 45^^\circ = \frac{AB}{BC} \implies 1 = \frac{100}{BC} \implies BC = 100 m$$

In right $\triangle ABD$:

$$\tan 30^^\circ = \frac{AB}{BD} \implies \frac{1}{\sqrt{3}} = \frac{100}{BD} \implies BD = 100\sqrt{3} m$$

Distance between the two ships $= CD = BC + BD$:

$$CD = 100 + 100\sqrt{3} = 100(1 + \sqrt{3}) = 100(1 + 1.732) = 100(2.732) = 273.2 m$$

Answer: $273.2 m$ (or $100(\sqrt{3}+1) m$)

🎯 Board Pattern (2018–2025): Case Study questions in Chapter 9 often involve a moving object (car/boat) or a change in elevation (like the boy walking towards the building). Always find the horizontal distances in both scenarios and subtract/add them.

📋 Board Revision Checklist

✅ Diagram check: Right angle marked? Observer level marked?
✅ Elevation = Looking UP; Depression = Looking DOWN
✅ If observer height is given (e.g., 1.5m boy), subtract it from the total object height
✅ $\tan 30^^\circ = 1/\sqrt{3} ≈ 0.577$
✅ $\tan 45^^\circ = 1$ (Shadow length = Object height)
✅ $\tan 60^^\circ = \sqrt{3} ≈ 1.732$
✅ For "broken tree" problems: Total height = Standing part + Hypotenuse (Broken part)
✅ Alternate angles: Depression from top = Elevation from ground

💡 Exam Tip:
In double-triangle problems, always solve the triangle with more known information first (usually the one with $45^^\circ$ angle as it makes Base = Perpendicular).

CBSE Section E • 4 Marks

Case Study: Marine Lighthouse Coastal Surveillance

Scenario: As observed from the top of a 75 m high lighthouse from sea level, the angles of depression of two ships sailing towards the foot of the lighthouse on the same side are 30° and 45° respectively.

Q1: State whether the nearer ship corresponds to the 45° angle of depression or the 30° angle. 1 Mark

Q2: Find the distance of the nearer ship from the foot of the lighthouse. 1 Mark

Q3: Calculate the exact distance between the two ships (take $\sqrt{3} \approx 1.732$). (OR) If both ships travel towards each other at equal speed, at what distance from the lighthouse will they cross? 2 Marks

📝 10-Mark Self-Assessment Test

Time: 25 Mins

Solve these high-frequency CBSE Board questions on paper, then reveal step-by-step solutions to check your marks.

Q1 (1 Mark - MCQ): The angle of elevation of the top of a tower from a point on the ground which is $30 m$ away from the foot of the tower is $30^^\circ$. The height of the tower is:
(a) $10\sqrt{3} m$ (b) $30\sqrt{3} m$ (c) $10 m$ (d) $20\sqrt{3} m$

Q2 (2 Marks - VSA): A kite is flying at a height of $60 m$ above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is $60^^\circ$. Find the length of the string, assuming that there is no slack in the string.

Q3 (3 Marks - SA): A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle $30^^\circ$ with it. The distance between the foot of the tree to the point where the top touches the ground is $8 m$. Find the total height of the tree.

Q4 (4 Marks - LA): From the top of a $7 m$ high building, the angle of elevation of the top of a cable tower is $60^^\circ$ and the angle of depression of its foot is $45^^\circ$. Determine the height of the tower.

Sol 1: $\tan 30^^\circ = \frac{h}{30} \implies \frac{1}{\sqrt{3}} = \frac{h}{30} \implies h = \frac{30}{\sqrt{3}} = 10\sqrt{3} m$.
Correct Option: (a) $10\sqrt{3} m$ [1 Mark]

Sol 2: Let length of string be $L$ (hypotenuse). Perpendicular height $= 60 m$.
$$\sin 60^^\circ = \frac{Perpendicular}{Hypotenuse} = \frac{60}{L} \implies \frac{\sqrt{3}}{2} = \frac{60}{L}$$ $$L = \frac{120}{\sqrt{3}} = \frac{120\sqrt{3}}{3} = 40\sqrt{3} m$$ Answer: Length of string $= 40\sqrt{3} m$ [2 Marks]

Sol 3: Let $AB$ be standing stem and $AC$ be broken hypotenuse. Base $BC = 8 m$, $\angle C = 30^^\circ$.
1. $\tan 30^^\circ = \frac{AB}{8} \implies \frac{1}{\sqrt{3}} = \frac{AB}{8} \implies AB = \frac{8}{\sqrt{3}} m$ [1 Mark]
2. $\cos 30^^\circ = \frac{BC}{AC} \implies \frac{\sqrt{3}}{2} = \frac{8}{AC} \implies AC = \frac{16}{\sqrt{3}} m$ [1 Mark]
Total height $= AB + AC = \frac{8}{\sqrt{3}} + \frac{16}{\sqrt{3}} = \frac{24}{\sqrt{3}} = 8\sqrt{3} m$ [1 Mark]
Answer: $8\sqrt{3} m$ [3 Marks]

Sol 4: Let $AB = 7 m$ be building and $CD$ be tower. Draw horizontal $AE \perp CD$. Then $ED = AB = 7 m$.
In right $\triangle ABD$: $\tan 45^^\circ = \frac{AB}{BD} \implies 1 = \frac{7}{BD} \implies BD = 7 m$.
Therefore $AE = BD = 7 m$ [1.5 Marks].
In right $\triangle CEA$: $\tan 60^^\circ = \frac{CE}{AE} \implies \sqrt{3} = \frac{CE}{7} \implies CE = 7\sqrt{3} m$ [1.5 Marks].
Total height of tower $= CE + ED = 7\sqrt{3} + 7 = 7(\sqrt{3} + 1) m$.
Answer: $7(\sqrt{3} + 1) m$ [4 Marks]

Concept Mastery Quiz 🎯

Test your readiness for the board exam.

1. The length of a shadow of a tower is $\sqrt{3}$ times its height. The angle of elevation of the sun is:

2. A bridge over a river makes an angle of $45^^\circ$ with the river bank. If the length of the bridge across the river is 150m, the width of the river is:

3. The angle of depression of a car parked on the road from the top of a 150m high tower is $30^^\circ$. The distance of the car from the tower is:

4. If a pole 6m high casts a shadow $2\sqrt{3}$m long on the ground, then the sun's elevation is:

5. A ladder 10m long reaches a window 8m above the ground. The distance of the foot of the ladder from the base of the wall is:

Frequently Asked Questions: Class 10 Maths Chapter 9

How many marks does Applications of Trigonometry carry in CBSE Class 10?

Chapter 9 contributes 4 to 5 marks in the board exam, almost always appearing as a 4-mark Case Study question or a 5-mark structured problem on heights and distances.

What is the difference between Angle of Elevation and Angle of Depression?

Angle of elevation is the acute angle measured upward from the horizontal line of sight to an object above. Angle of depression is measured downward from the horizontal line of sight to an object below. By alternate interior angles, angle of depression equals angle of elevation from the object to the observer.

Which trigonometric ratios are most commonly used in Heights and Distances?

tan(theta) = Opposite / Adjacent is used in over 80% of problems involving heights and horizontal distances. sin(theta) and cos(theta) are used when ladder lengths, kite strings, or inclined slopes are involved.

Why is drawing a diagram essential in height and distance problems?

In CBSE marking schemes, 1 to 1.5 marks are allocated strictly for drawing a neat, correctly labelled right-triangle diagram with correct angle placement.

Chapter 9 Study Modules & Practice Resources

Complete Notes & Formulas • NCERT Solutions • Previous Year Questions • Chapter Tests & Quizzes • Practice Worksheets

Chapter 9: Applications of Trigonometry

Overview