What are the standard numerical values of $\sqrt{3}$ and $\sqrt{2}$ used in heights and distances?

Unless otherwise instructed: \sqrt{3} \approx 1.732 and \sqrt{2} \approx 1.414. If the question does not explicitly specify substituting decimal values, leave the answer in exact radical form (e.g., 20(\sqrt{3}-1)\text{ m}) after rationalising denominators.

How do you solve problems where a tower subtends angles of elevation from two points on opposite sides of the base?

Draw two right triangles with a shared common vertical tower height h. Let distances from the base be x and y. Express \tan\theta_1 = \frac{h}{x} \implies x = \frac{h}{\tan\theta_1} and \tan\theta_2 = \frac{h}{y} \implies y = \frac{h}{\tan\theta_2}. The total distance between points is x + y = h(\cot\theta_1 + \cot\theta_2).

What is the common mistake in questions involving an observer with a given height (e.g. 1.5 m boy)?

Students often calculate the total tower height from the ground directly. If the observer has height 1.5\text{ m}, the right triangle calculates the vertical distance above eye level h'. The total height of the monument is H = h' + 1.5\text{ m}.

How do you approach a multi-stage problem like a flagstaff on top of a building?

Draw two right triangles sharing the same horizontal distance x from the base. Let building height be h and flagstaff height be p. Set \tan\alpha = \frac{h}{x} and \tan\beta = \frac{h + p}{x}. Equate x = \frac{h}{\tan\alpha} = \frac{h+p}{\tan\beta} to solve for h or p.

Why should denominators always be rationalised in final heights and distances solutions?

Expressions like \frac{60}{\sqrt{3}} should be converted to \frac{60\sqrt{3}}{3} = 20\sqrt{3}\text{ m}. Rationalising simplifies arithmetic evaluation and prevents precision loss during decimal substitution.

How many right triangles are involved in CBSE Class 10 heights and distances questions?

CBSE curriculum restricts problems to at most TWO right-angled triangles with standard angles (30^\circ, 45^\circ, 60^\circ). Problems requiring three-dimensional visualization or non-standard angles are excluded.

All Revision Notes

MathematicsCh-9 15 min comprehensive revision

NCERT Class 10 Mathematics — Chapter 9

Some Applications of Trigonometry

Line of sight, Angle of Elevation, Angle of Depression, two-building observations, lighthouse ship tracking, and multi-triangle height-and-distance problems.

Quick Key Takeaways:

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

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 (always equals angle of elevation by alternate interior angles).

Key Radical Approximations:

\sqrt{3} \approx 1.732

and

\sqrt{2} \approx 1.414

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Jump to Concept:1. 1. Line of Sight, Elevation vs Depression & Geometric Modeling 2. 2. Solving Multi-Triangle Systems (Two Heights / Two Angles)3. 3. Important Solved Board Examination Questions (3-Mark & 5-Mark)4. 4. CBSE Case-Study Modeling: Lighthouse Maritime Navigation & Distress Signal 5. 5. CBSE Examiner Marking Scheme, Calculation Traps & Step Rubric

1. Line of Sight, Elevation vs Depression & Geometric Modeling

Axioms & Foundational Theory

Comprehensive mathematical foundations, definitions, formulas, and visual conceptual model for Some Applications of Trigonometry.

• Geometry of Elevation and Depression Angles

•

Horizontal Reference Line: Every angle of elevation or depression MUST be measured from a horizontal line drawn at the observer's eye level.

•

Alternate Interior Angles Equivalence: Since the horizontal line at eye level is parallel to the ground horizontal, the Angle of Depression from an elevated observer equals the Angle of Elevation from the ground target.

📊 Some Applications of Trigonometry: Heights & DistancesVisual Model

Visual schematic mapping the observer line-of-sight, angle of elevation, angle of depression, and the two-triangle trigonometric height formula.

2. Solving Multi-Triangle Systems (Two Heights / Two Angles)

Theorem Proofs & Derivations

Step-by-step rigorous geometric and algebraic theorem proofs and formula derivations for Some Applications of Trigonometry.

• Two-Triangle System Solution Strategy

•

Step 1: Draw a neat, fully labeled geometric diagram showing all vertical heights (

h

), horizontal distances (

x

), and angles of elevation/depression.

•

Step 2: Identify the common side (usually the horizontal distance

x

between bases) connecting the two right-angled triangles.

•

Step 3: Set up

\tan\theta_1 = \frac{h_1}{x}

and

\tan\theta_2 = \frac{h_2}{x}

•

Step 4: Eliminate

x

by equating expressions and solve for the unknown height

h

3. Important Solved Board Examination Questions (3-Mark & 5-Mark)

Topper Step Solutions

Standard CBSE board exam numericals and riders with complete step-by-step mathematical working and justifications.

• 3-Mark Standard Board Question: From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30° and 45°, respectively. If the bridge is at a height of 3 m from the banks, find the width of the river.

•

Step 1 (Diagram & Variables):
Let

P

be the point on the bridge at height

PQ = 3

m above the river bed.
Let

A

and

B

be points on opposite banks.
In right

\Delta PQA

\angle PAQ = 30^\circ

. In right

\Delta PQB

\angle PBQ = 45^\circ

•

Step 2 (In $\Delta PQA$ ):

\tan 30^\circ = \frac{PQ}{AQ} \implies \frac{1}{\sqrt{3}} = \frac{3}{AQ} \implies AQ = 3\sqrt{3}\text{ m}

•

Step 3 (In $\Delta PQB$ ):

\tan 45^\circ = \frac{PQ}{QB} \implies 1 = \frac{3}{QB} \implies QB = 3\text{ m}

•

Step 4 (Total Width of River $AB$ ):

AB = AQ + QB = 3\sqrt{3} + 3 = 3(\sqrt{3} + 1)\text{ m}

AB = 3(1.732 + 1) = 3(2.732) = 8.196\text{ m}

•

Final Boxed Answer:

\mathbf{\text{Width of River} = 3(\sqrt{3} + 1)\text{ m} \approx 8.2\text{ m}}

• 5-Mark Heavyweight Board Problem / Rider: A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30°. Find the distance travelled by the balloon during the interval.

•

Step 1 (Effective Vertical Height):
Total height of balloon

= 88.2

m. Height of girl

= 1.2

m.
Effective height above eye level

h = 88.2 - 1.2 = \mathbf{87\text{ m}}

•

Step 2 (First Position of Balloon - Angle $60^\circ$ ):
Let horizontal distance be

x_1

\tan 60^\circ = \frac{h}{x_1} \implies \sqrt{3} = \frac{87}{x_1} \implies x_1 = \frac{87}{\sqrt{3}} = \frac{87\sqrt{3}}{3} = 29\sqrt{3}\text{ m}

•

Step 3 (Second Position of Balloon - Angle $30^\circ$ ):
Let horizontal distance be

x_2

\tan 30^\circ = \frac{h}{x_2} \implies \frac{1}{\sqrt{3}} = \frac{87}{x_2} \implies x_2 = 87\sqrt{3}\text{ m}

•

Step 4 (Distance Travelled $d = x_2 - x_1$ ):

d = 87\sqrt{3} - 29\sqrt{3} = (87 - 29)\sqrt{3} = \mathbf{58\sqrt{3}\text{ m}}

d = 58 \times 1.732 = 100.456\text{ m}

•

Final Boxed Answer:

\mathbf{\text{Distance Travelled} = 58\sqrt{3}\text{ m} \approx 100.46\text{ m}}

4. CBSE Case-Study Modeling: Lighthouse Maritime Navigation & Distress Signal

Case Study Mastery (4 Marks)

Real-world mathematical modeling scenario with structured multi-part questions and full step solutions.

• Practical Application Context: Lighthouse Maritime Navigation & Distress Signal

As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

•

Q1: What are the angles of elevation from the ships to the top of the lighthouse?

\rightarrow

By alternate interior angles, the angles of elevation are

45^\circ

(nearer ship) and

30^\circ

(farther ship).

•

Q2: Calculate the distance of the nearer ship from the base of the lighthouse.

\rightarrow

\Delta AB C

\tan 45^\circ = \frac{75}{d_1} \implies 1 = \frac{75}{d_1} \implies d_1 = \mathbf{75\text{ meters}}

•

Q3: Calculate the distance between the two ships.

\rightarrow

\Delta ABD

\tan 30^\circ = \frac{75}{d_2} \implies \frac{1}{\sqrt{3}} = \frac{75}{d_2} \implies d_2 = 75\sqrt{3}

m. Distance between ships

= d_2 - d_1 = 75\sqrt{3} - 75 = 75(\sqrt{3} - 1) = 75(1.732 - 1) = 75(0.732) = \mathbf{54.9\text{ meters}}

5. CBSE Examiner Marking Scheme, Calculation Traps & Step Rubric

Important Solved Board Questions

Examiner step-marking allocations, common algebraic/arithmetic deduction traps, and presentation guidelines.

• Step-by-Step Marking Rubric & Presentation Guidelines

•

1 Mark: Neat, accurate geometric diagram with all angles and heights labeled.

•

1 Mark: Subtracting observer height to find effective height (e.g.

88.2 - 1.2 = 87

•

2 Marks: Setting up correct

\tan\theta

ratios in both triangles and solving for horizontal components.

•

1 Mark: Final rationalized answer with units (meters).

• Common Calculation Traps & Verification Checklist

•

Trap 1: Forgetting to draw a horizontal line at the top when representing angles of depression.

•

Trap 2: Forgetting to subtract the observer's eye-height from the total height.

•

Trap 3: Leaving radical in the denominator (always rationalize

\frac{87}{\sqrt{3}} = 29\sqrt{3}

Authentic Board Question (3 Marks)Topic: Some Applications of Trigonometry Mathematical Proof & Computations

State the governing mathematical theorem/formula, show complete geometric or computational steps, and find the exact result for Some Applications of Trigonometry.

Official CBSE Step-by-Step Marking Breakdown:

Step 1: Given / To Prove / Formula Setup: State the governing theorem (BPT, Tangents, Pythagoras), mensuration formula, or trigonometric ratios with given values.

1 Mark

Step 2: Step-by-Step Derivation / Arithmetic Working: Show complete deductive steps or numerical calculations with

\pi = 22/7

or trigonometric standard values.

1 Mark

Step 3: Final Boxed Answer with Units / Q.E.D.: State the final numerical result with proper square/cubic units (

\text{cm}^2, \text{m}^3

) or formal proof conclusion.

1 Mark

Model Student Answer (Target: Full 3/3 Marks):

To score full 3 marks on Some Applications of Trigonometry in CBSE Mathematics:

1. Theorem / Formula: Write the standard governing formula or state the geometric theorem clearly.
2. Calculation / Proof: Substitute dimensions systematically or provide deductive step justifications.
3. Final Result: Box the final answer with required units (

\text{cm}, \text{cm}^2, \text{m}^3

) or conclude with "Hence Proved".

Examiner Mark Deduction Traps:

•Always write intermediate calculation lines to earn partial credit under CBSE step-marking.

•State reasons in parentheses (e.g., "[Tangents from an external point are equal]") during geometric proofs.

High-Frequency Conceptual Doubts & FAQs

Curated answers to the most common questions asked by Class 10 students.

Angle of Elevation is measured upward from the observer's horizontal line of sight to an object placed above. Angle of Depression is measured downward from the horizontal line of sight to an object placed below. By alternate interior angles, the Angle of Depression from an observer equals the Angle of Elevation from the object.

Some Applications of Trigonometry

1. Line of Sight, Elevation vs Depression & Geometric Modeling

2. Solving Multi-Triangle Systems (Two Heights / Two Angles)

3. Important Solved Board Examination Questions (3-Mark & 5-Mark)

4. CBSE Case-Study Modeling: Lighthouse Maritime Navigation & Distress Signal

5. CBSE Examiner Marking Scheme, Calculation Traps & Step Rubric

Official CBSE Step-by-Step Marking Breakdown:

High-Frequency Conceptual Doubts & FAQs

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