Complete Solutions and Summary of Some Applications of Trigonometry – NCERT Class 10, Mathematics, Chapter 9 – Summary, Questions, Answers, Extra Questions

Comprehensive summary and explanation of Chapter 9 'Some Applications of Trigonometry', covering concepts of angles of elevation and depression, line of sight, application of trigonometric ratios in calculating heights and distances in real-life problems, worked examples with step-by-step solutions, and all related NCERT questions and extra practice exercises.

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Categories: NCERT, Class X, Mathematics, Summary, Extra Questions, Trigonometry, Applications, Heights and Distances, Real-life Problems, Chapter 9

Tags: Trigonometry, Heights and Distances, Angle of Elevation, Angle of Depression, Line of Sight, Real-life Applications, Problem Solving, NCERT, Class 10, Mathematics, Chapter 9, Answers, Extra Questions

Some Applications of Trigonometry Class 10 NCERT Chapter 9 - Ultimate Study Guide, Notes, Questions, Quiz 2025

Full Chapter Summary & Detailed Notes - Some Applications of Trigonometry Class 10 NCERT

Overview & Key Concepts

Chapter Goal: Apply trigonometric ratios to find heights and distances in real-life scenarios. Exam Focus: Angles of elevation/depression, word problems on towers, buildings, rivers. 2025 Updates: Emphasis on practical applications like surveying. Fun Fact: Trigonometry used in ancient architecture like pyramids. Core Idea: Use right triangles formed by line of sight. Real-World: Measuring tall structures without climbing.
Wider Scope: Foundation for physics (projectiles), engineering (construction), navigation.

9.1 Heights and Distances

Builds on Chapter 8 trig ratios; applies to everyday problems.
Line of Sight: Line from observer's eye to object point.
Angle of Elevation: Formed by line of sight with horizontal when object above (raise head). Example: Student looking at minar top, angle BAC.
Angle of Depression: Formed when object below (lower head). Example: Girl on balcony looking at flower pot, angle below horizontal.
Process: Identify right triangle, known angles/distances, use tan/sin/cos to find unknown.
Information Needed: Distance from object, angle of elevation/depression, observer height.
Example Scenarios: Towers, poles, chimneys, flags, shadows, buildings, rivers, bridges.

Example 1: Tower Height with 60° Elevation

Tower AB vertical, point C 15m away, angle ACB=60°. Right triangle ABC at B.
tan60°=AB/BC → √3=AB/15 → AB=15√3 m.
Detailed: Use opposite/adjacent; verify with sin/cos if needed.

Example 2: Ladder for Pole Repair

Pole AD=5m, repair at B=1.3m below top, BD=3.7m. Ladder BC at 60° to horizontal.
sin60°=BD/BC → (√3/2)=3.7/BC → BC=3.7*(2/√3)≈4.28m.
cot60°=DC/BD → (1/√3)=DC/3.7 → DC=3.7/√3≈2.14m.
Detailed: Right triangle BDC at D; hypotenuse ladder; base distance from pole.

Example 3: Chimney Height with Observer

Observer CD=1.5m tall, 28.5m from chimney AB, angle ADE=45°.
tan45°=AE/DE → 1=AE/28.5 → AE=28.5m.
AB=AE+BE=28.5+1.5=30m.
Detailed: Triangle ADE right at E; BE=observer height.

Example 4: Flag on Building

Building AB=10m, flag BD=x, point P, angle 30° to AB top, 45° to AD top.
tan30°=AB/PA → 1/√3=10/PA → PA=10√3 m.
tan45°=AD/PA → 1=(10+x)/(10√3) → x=10(√3-1)≈7.32m.
Detailed: Triangles PAB (30°), PAD (45°); same base PA.

Example 5: Tower Shadow

Tower AB=h, shadow BC=x at 60°, DB=40+x at 30°.
tan60°=h/x → √3=h/x → h=x√3.
tan30°=h/(40+x) → 1/√3=h/(40+x) → x√3/(40+x)=1/√3 → 3x=x+40 → x=20, h=20√3 m.
Detailed: Two triangles ABC (60°), ABD (30°); relate shadows.

Example 6: Multi-Storeyed Building Depression

Building PC, 8m building AB; angles 30° to AB top, 45° to bottom from PC top.
PB transversal, angles equal: PBD=30°, PAC=45°.
tan30°=PD/BD → BD=PD√3.
tan45°=PC/AC → PC=AC.
PC=PD+8, AC=BD → PD+8=PD√3 → PD=8/(√3-1)=4(√3+1) m.
Height PC=4(√3+1)+8=4√3+12 m; distance 4(√3+1)√3=4(3+√3) m.
Detailed: Parallel lines PQ/BD; alternate angles; equate distances.

Example 7: Bridge Across River

Bridge PD=3m high; angles 30° to A, 45° to B opposite banks.
AB=AD+DB; tan30°=PD/AD → 1/√3=3/AD → AD=3√3 m.
tan45°=PD/BD → 1=3/BD → BD=3 m.
Width AB=3+3√3=3(1+√3) m.
Detailed: Triangles APD (30°), BPD (45°); same height PD.

Exercise 9.1

15 problems: Circus rope, broken tree, slides, tower height, kite, boy walking, transmission tower, statue, building-tower angles, poles on road, TV tower canal, cable tower, lighthouse ships, balloon, highway car.
Types: Single angle, two angles, heights with observers, shadows, depressions.

9.2 Summary

Line of sight, elevation (raise head), depression (lower head).
Use trig to find heights/distances.

Why This Guide Stands Out

Complete chapter coverage: Notes, examples, Q&A (all NCERT + extras), quiz. Student-centric, exam-ready for 2025. Free & ad-free.

Key Themes & Tips

Concepts: Elevation/depression angles, right triangles.
Methods: Tan for height/distance, sin/cos for hypotenuse.
Problems: Account for observer height, multiple triangles.
Tip: Draw diagrams; use positive values; approximate if needed (√3≈1.732).

Exam Case Studies

Word problems on inaccessible heights/distances; nature of solutions (positive).

Project & Group Ideas

Measure school building height using clinometer; model river width.

Key Definitions & Terms - Complete Glossary

All terms from chapter; easy to memorize. Detailed explanations with examples for better understanding.

Line of Sight

The straight line drawn from the eye of an observer to the point on the object being viewed. Example: From student's eye to minar top. It forms the hypotenuse in the right triangle used for calculations. In problems, it's crucial for identifying the angle with horizontal.

Angle of Elevation

The angle formed by the line of sight with the horizontal when the object is above the observer's level (raise head to see). Example: Looking at tower top from ground, angle BAC=60°. Measured from horizontal up; used with tan = opposite/adjacent for height.

Angle of Depression

The angle formed by the line of sight with the horizontal when the object is below the observer's level (lower head to see). Example: From balcony to flower pot, angle below horizontal. Equal to elevation from object's view; used in multi-level problems.

Horizontal Level

The imaginary flat line at the observer's eye height, parallel to ground. Example: In figures, line AE or similar. Reference for measuring elevation/depression angles; observer height added/subtracted as needed.

Right Triangle in Applications

Triangle formed by observer, object, and foot, right-angled at base. Example: ABC with right angle at B. Trig ratios applied: opposite (height), adjacent (distance), hypotenuse (line of sight).

Observer Height

The height of the observer's eye from ground, added to calculated height. Example: 1.5m in chimney problem. Ignored if not mentioned; key in accurate measurements.

Tip: Use flashcards; visualize with diagrams. Remember: Elevation up, depression down; both use same trig principles.

6. tan30° value?

1 Mark Answer: 1/√3.

7. tan45° value?

1 Mark Answer: 1.

8. sin60° value?

1 Mark Answer: √3/2.

9. cot60° value?

1 Mark Answer: 1/√3.

10. Tower 15m away, 60° elevation height?

1 Mark Answer: 15√3 m.

11. Ladder length for 3.7m at 60°?

1 Mark Answer: 4.28m.

12. Chimney height with 1.5m observer?

1 Mark Answer: 30m.

13. Flag length on 10m building?

1 Mark Answer: 7.32m.

14. Tower shadow 40m longer?

1 Mark Answer: 20√3 m.

15. Multi-building height?

1 Mark Answer: 4(√3+3) m.

16. River width with 3m bridge?

1 Mark Answer: 3(1+√3) m.

17. Rope height with 30°?

1 Mark Answer: 10√3 m.

18. Broken tree height?

1 Mark Answer: 8(√3+1) m.

19. Slide length 1.5m 30°?

1 Mark Answer: 3m.

20. Kite string 60m 60°?

1 Mark Answer: 60m.

Part B: 4 Marks Questions (Answers in ~6 Lines)

1. Explain line of sight with example.

4 Marks Answer: Line from observer eye to object point. In minar fig, AC from student eye to top. Forms right triangle with horizontal. Used to define elevation angle BAC.

2. Difference elevation vs depression.

4 Marks Answer: Elevation when object above, raise head; depression below, lower head. Both angles with horizontal; trig ratios same.

3. Information for minar height.

4 Marks Answer: Distance DE, angle BAC, student height AE. CD=CB+BD, BD=AE, tanA=BC/AB.

4. Tower 15m away 60° height.

4 Marks Answer: tan60°=AB/15=√3, AB=15√3 m. Right triangle at B.

5. Ladder for 3.7m at 60°.

4 Marks Answer: sin60°=3.7/BC, BC=3.7*2/√3≈4.28m. cot60°=DC/3.7, DC=3.7/√3≈2.14m.

6. Chimney with 1.5m observer 45°.

4 Marks Answer: tan45°=AE/28.5=1, AE=28.5m. AB=30m.

7. Flag on 10m building 30°/45°.

4 Marks Answer: PA=10√3. tan45°=(10+x)/PA=1, x=10(√3-1)≈7.32m.

8. Shadow 40m longer 30°/60°.

4 Marks Answer: h=x√3, h=(x+40)/√3, x=20, h=20√3 m.

9. Multi-building 30°/45° 8m.

4 Marks Answer: BD=PD√3, PC=AC, PD+8=PD√3, PD=8/(√3-1)=4(√3+1)m. Height 4√3+12m.

10. River 3m bridge 30°/45°.

4 Marks Answer: AD=3√3, BD=3, AB=3(1+√3)m.

11. Rope 20m 30° height.

4 Marks Answer: sin30°=h/20=1/2, h=10m.

12. Tree broken 30° 8m.

4 Marks Answer: tan30°=h/8=1/√3, h=8/√3. Total 8/√3 +8=8(1+1/√3).

13. Slide 1.5m 30° length.

4 Marks Answer: sin30°=1.5/l=1/2, l=3m.

14. Tower 30m away 30° height.

4 Marks Answer: tan30°=h/30=1/√3, h=10√3 m.

15. Kite 60m height 60° string.

4 Marks Answer: sin60°=60/l=√3/2, l=120/√3=40√3 m.

16. Boy walks 30° to 60° distance.

4 Marks Answer: d1= (30-1.5)√3, d2=(30-1.5)/√3, walked d1-d2.

17. Transmission tower height.

4 Marks Answer: tan45°=20/d=1, d=20. tan60°=(20+h)/20=√3, h=20(√3-1).

18. Statue on pedestal 60°/45°.

4 Marks Answer: tan60°=(h+1.6)/d=√3, tan45°=h/d=1. h=(1.6√3)/(√3-1).

19. Building tower 30°/60° 50m.

4 Marks Answer: tan60°=50/d=√3, d=50/√3. tan30°=h/d=1/√3, h=50/3 m.

20. Poles equal 60°/30° 80m road.

4 Marks Answer: tan60°=h/x=√3, tan30°=h/(80-x)=1/√3. x=60, h=20√3 m.

Part C: 8 Marks Questions (Detailed Answers)

1. Explain angle of elevation with diagram.

8 Marks Answer: When object above horizontal, line of sight makes angle with horizontal called elevation. In fig 9.1, student at A, minar CD, eye E, line AC to top C. Horizontal AE, angle BAC elevation. Right triangle ABC at B, tanA = BC/AB. Add observer height AE to BC for total CD. Used when raising head. Example: Tower view from ground. Key: Identifies opposite (height), adjacent (distance).

2. Explain angle of depression with diagram.

8 Marks Answer: When object below horizontal, angle below called depression. In fig 9.3, girl on balcony, flower pot below, line of sight down. Horizontal level, angle with it. Right triangle, tan = opposite/adjacent. Equal to elevation from pot's view. Example: From building to ground object. Key: Lower head; often in multi-level like bridges, buildings.

3. Process to find minar height.

8 Marks Answer: Need DE (distance), angle BAC (elevation), AE (student height). CD = CB + BD, BD=AE. In triangle ABC right at B, tanA = BC/AB, BC = AB tanA. Total height BC + AE. If cotA, AB = BC cotA. Choose ratio with known/unknown. Example: If AB=15m, A=60°, BC=15√3, add AE.

4. Solve tower 15m 60°.

8 Marks Answer: AB tower, CB=15m, angle ACB=60°. Right at B. tan60° = AB/15 = √3. AB=15√3 m. Verify sin60° = AB/AC, AC=AB/(√3/2)=30/√3=10√3, cos60°=15/AC=1/2, same. No observer height, assume ground level.

5. Solve ladder pole repair.

8 Marks Answer: Pole AD=5m, B=1.3m below, BD=3.7m. Ladder BC at 60°. Right BDC at D. sin60°=3.7/BC=√3/2, BC=3.7*2/√3 = (7.4/√3)* (√3/√3)=7.4√3/3≈4.28m. cot60°=DC/3.7=1/√3, DC=3.7/√3≈2.14m. Use √3≈1.73. Verify tan60°=3.7/DC=√3, same.

6. Solve chimney observer.

8 Marks Answer: Chimney AB, observer CD=1.5m, DE=28.5m, angle ADE=45°. Right at E. tan45°=AE/DE=1, AE=28.5m. AB=AE+BE, BE=1.5m (observer height), AB=30m. Diagram: Eye D, horizontal DE, line DA to top. AE is calculated height above eye, add eye height.

7. Solve flag building.

8 Marks Answer: Building AB=10m, flag BD=x, AD=10+x. Point P, angle PAB=30°, PAD=45°. tan30°=10/PA=1/√3, PA=10√3≈17.32m. tan45°=(10+x)/PA=1, 10+x=10√3, x=10(√3-1)≈7.32m. Two triangles PAB, PAD sharing PA. Distance PA same.

8. Solve shadow tower.

8 Marks Answer: Tower AB=h, shadow BC=x at 60°, DB=x+40 at 30°. tan60°=h/x=√3, h=x√3. tan30°=h/(x+40)=1/√3. Substitute h: x√3 /(x+40)=1/√3, 3x =x+40, 2x=40, x=20, h=20√3 m. Two scenarios for sun angles; longer shadow at lower angle.

9. Solve multi-building.

8 Marks Answer: Multi PC, 8m AB. From P, depression 30° to B, 45° to A. PB transversal, angles PBD=30°, PAC=45°. tan30°=PD/BD=1/√3, BD=PD√3. tan45°=PC/AC=1, PC=AC. PC=PD+DC, DC=8. AC=BD, PD+8=PD√3. PD(√3-1)=8, PD=8/(√3-1)=4(√3+1) m. PC=4(√3+1)+8=4√3+12 m. Distance AC=4(√3+1)√3=12+4√3 m.

10. Solve river bridge.

8 Marks Answer: Bridge P height PD=3m, angles 30° to A, 45° to B. AB=AD+DB. tan30°=3/AD=1/√3, AD=3√3 m. tan45°=3/BD=1, BD=3 m. AB=3+3√3=3(1+√3) m. Two triangles APD, BPD sharing height; opposite banks.

11. Solve circus rope.

8 Marks Answer: Rope 20m, angle 30° with ground. Right triangle, sin30°=h/20=1/2, h=10m. Or tan30°=h/d=1/√3, d=10√3 m, hypotenuse 20m verify.

12. Solve broken tree.

8 Marks Answer: Broken part bends 30° to ground, touch point 8m from foot. Let height h broken, total H=h + unbroken. tan30°=h/8=1/√3, h=8/√3. Unbroken=8m (hypotenuse? No, bent like hypotenuse). Broken part hypotenuse l, sin30°=8/l, l=16m, cos30°=unbroken/l, unbroken=16*(√3/2)=8√3 m. Total h + unbroken? Book: Distance 8m to touch, angle 30°, height unbroken + broken. Let unbroken a, broken b, tan30°=a/8, a=8/√3. b hypotenuse, b=8 / sin30°=16m. Total a+b=8/√3 +16=8(2 +1/√3)=8(√3+1)/√3 *√3/√3=8(3+√3)/3.

13. Solve slides contractor.

8 Marks Answer: Small slide 1.5m high, 30°. sin30°=1.5/l=1/2, l=3m. Steep 3m high, 60°. sin60°=3/l=√3/2, l=3*2/√3=2√3 m. Two cases, different angles.

14. Solve tower 30m 30°.

8 Marks Answer: tan30°=h/30=1/√3, h=30/√3=10√3 m. Simple single angle.

15. Solve kite 60m 60°.

8 Marks Answer: Height 60m above ground, angle 60°. sin60°=60/l=√3/2, l=120/√3=40√3 m. String is hypotenuse.

16. Solve boy walking building.

8 Marks Answer: Boy 1.5m tall, building 30m. Angle from eyes 30° to 60° as walks. Let distance d1 at 30°, d2 at 60°. tan30°=(30-1.5)/d1=1/√3, d1=28.5√3. tan60°=28.5/d2=√3, d2=28.5/√3. Walked d1-d2=28.5(√3 -1/√3)=28.5*(2/√3)=19√3 m approx.

17. Solve transmission tower.

8 Marks Answer: Building 20m, tower h on top. Angles 45° bottom, 60° top. tan45°=20/d=1, d=20m. tan60°=(20+h)/20=√3, 20+h=20√3, h=20(√3-1) m.

18. Solve statue pedestal.

8 Marks Answer: Statue 1.6m on pedestal h. Angle top statue 60°, top pedestal 45°. tan60°=(h+1.6)/d=√3, tan45°=h/d=1, d=h. h+1.6 = h√3, 1.6=h(√3-1), h=1.6/(√3-1)=0.8(√3+1) m.

19. Solve building tower angles.

8 Marks Answer: Building h, tower 50m. Angle building to tower 60°, tower to building 30°. tan60°=50/d=√3, d=50/√3. tan30°=h/d=1/√3, h=d/√3=50/3 m.

20. Solve poles road.

8 Marks Answer: Poles h equal, road 80m. Point between, angles 60° one, 30° other. Let distance x to 60°, 80-x to 30°. tan60°=h/x=√3, h=x√3. tan30°=h/(80-x)=1/√3. x√3 /(80-x)=1/√3, 3x=80-x, 4x=80, x=20. h=20√3 m. Distances 20m, 60m.

Practice Tip: Draw diagrams; use trig table; approximate for checks.

All Textbook Examples - Step by Step Solutions

Complete list with detailed explanations like the book, step by step, including math and reasoning.

Example 1: Tower Height

Tower AB vertical on ground. Point C 15m from foot B, angle elevation ACB=60°. Find AB. Diagram fig 9.4. Right triangle ABC at B. Choose tan60° as opp AB, adj BC known. tan60°=AB/BC=√3=AB/15. AB=15√3 m. Reasoning: Direct application, no observer height.

Example 2: Electrician Ladder

Pole AD=5m, fault at B 1.3m below top, BD=3.7m. Ladder BC at 60° to horizontal. Find BC length, DC distance. Diagram fig 9.5. Right BDC at D. Hyp BC ladder. sin60°=opp BD/hyp BC=√3/2=3.7/BC, BC=3.7*2/√3≈4.28m (use √3=1.73). cot60°=adj DC/opp BD=1/√3=DC/3.7, DC=3.7/√3≈2.14m. Reasoning: sin for hypotenuse, cot for base.

Example 3: Chimney Height

Observer CD=1.5m tall, DE=28.5m from chimney AB, angle ADE=45° from eyes D to top A. Find AB. Diagram fig 9.6. Right ADE at E. tan45°=AE/DE=1=AE/28.5, AE=28.5m. BE=CD=1.5m (eye height). AB=AE+BE=30m. Reasoning: Add observer height to calculated above-eye height.

Example 4: Flag on Building

Building AB=10m tall, flag BD hoisted top. From P, elevation to AB top 30°, to flag top 45°. Find BD, PA. Diagram fig 9.7. Right PAB at B, tan30°=10/PA=1/√3, PA=10√3 m≈17.32m. Right PAD at A? Wait, at foot? Actually right at A for both? No, triangles PAB (right at B? Book at A? Wait, book: Right at B for PAB, but fig shows P ground, A foot building. tan30°=AB/PA, yes right at A actually? Book: Right at A? Text: Right Δ PAB, PAD at A. tan30°=AB/PA, yes. tan45°=AD/PA, AD=10+x=PA=10√3, x=10(√3-1)≈7.32m. Reasoning: Same adjacent PA for two opposites.

Example 5: Tower Shadow

Tower AB=h on level ground. Shadow 40m longer when sun 30° than 60°. Find h. Diagram fig 9.8. At 60°, shadow BC=x, tan60°=h/x=√3, h=x√3. At 30°, DB=x+40, tan30°=h/(x+40)=1/√3. Substitute h: x√3 /(x+40)=1/√3, multiply both √3: 3x /(x+40)=1, 3x=x+40, 2x=40, x=20, h=20√3 m. Reasoning: Two sun positions, relate shadows.

Example 6: Multi-Storeyed Depression

Multi PC, 8m tall AB. From P top PC, depression 30° to B top AB, 45° to A bottom. Find PC, AC. Diagram fig 9.9. PB transversal to parallels PQ, BD. Alternate angles: PBD=QPB=30°, PAC=45°. Right PBD at D? Text: Right PBD, tan30°=PD/BD=1/√3, BD=PD√3. Right PAC at C? tan45°=PC/AC=1, PC=AC. PC=PD+DC, DC=AB=8m. AC=BD, PD+8=PD√3. PD(√3-1)=8, PD=8/(√3-1)* (√3+1)/(√3+1)=8(√3+1)/2=4(√3+1)m. PC=4(√3+1)+8=4√3+12 m. AC=4(√3+1)√3=4(3+√3)=12+4√3 m. Reasoning: Use parallels for angle equality, equate distances.

Example 7: Bridge River

From bridge P height 3m, depression 30°/45° to opposite banks A/B. Find width AB. Diagram fig 9.10. Right APD at D, tan30°=PD/AD=1/√3=3/AD, AD=3√3 m. Right BPD at D, tan45°=3/BD=1, BD=3 m. AB=AD+BD=3√3+3=3(√3+1) m. Reasoning: Same height, different angles to banks; add distances.

Tip: Always include approximations if given (√3=1.732); verify with alternate ratios.

Quick Revision Notes & Mnemonics - Exam-Ready

Organized by subtopics for easy revision; key points, formulas, tips per section.

Introduction & Concepts

Apply trig ratios to heights/distances.
Line of sight: Eye to object.
Elevation: Above horizontal.
Depression: Below horizontal.

Key Ratios & Values

tan theta = h/d.
30°: tan=1/√3, sin=1/2.
45°: tan=1, sin=1/√2.
60°: tan=√3, sin=√3/2.

Problem-Solving Process

Draw diagram, mark triangle.
Identify known: Angle, distance, height.
Use tan/sin/cos.
Add observer height.

Single Angle Problems

Towers: h = d tan theta.
Ladders: l = h / sin theta.
Observer: Total = calc + eye.

Two Angles/Levels

Shadows: h = x tan high = (x+l) tan low.
Buildings: Use parallels, equate.
Rivers: Add distances from angles.

Exercise Types

Ropes, trees, slides.
Towers, kites, boys walking.
Poles, TV towers, lighthouses.
Balloons, highways.

Mnemonics: SOH-CAH-TOA (Sin Opp/Hyp, Cos Adj/Hyp, Tan Opp/Adj). Elevation Up, Depression Down (EUDD).

Tip: Revise diagrams; practice 1 per type daily.

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Complete Solutions and Summary of Some Applications of Trigonometry – NCERT Class 10, Mathematics, Chapter 9 – Summary, Questions, Answers, Extra Questions

Some Applications of Trigonometry

Full Chapter Summary & Detailed Notes - Some Applications of Trigonometry Class 10 NCERT

Overview & Key Concepts

9.1 Heights and Distances

Example 1: Tower Height with 60° Elevation

Example 2: Ladder for Pole Repair

Example 3: Chimney Height with Observer

Example 4: Flag on Building

Example 5: Tower Shadow

Example 6: Multi-Storeyed Building Depression

Example 7: Bridge Across River

Exercise 9.1

9.2 Summary

Why This Guide Stands Out

Key Themes & Tips

Exam Case Studies

Project & Group Ideas

Key Definitions & Terms - Complete Glossary

Line of Sight

Angle of Elevation

Angle of Depression

Horizontal Level

Right Triangle in Applications

Observer Height

60+ Questions & Answers - NCERT Based (Class 10)

Part A: 1 Mark Questions (Short Answers)

1. What is line of sight?

2. Define angle of elevation.

3. Define angle of depression.

4. Ratio for height/distance?

5. tan60° value?

6. tan30° value?

7. tan45° value?

8. sin60° value?

9. cot60° value?

10. Tower 15m away, 60° elevation height?

11. Ladder length for 3.7m at 60°?

12. Chimney height with 1.5m observer?

13. Flag length on 10m building?

14. Tower shadow 40m longer?

15. Multi-building height?

16. River width with 3m bridge?

17. Rope height with 30°?

18. Broken tree height?

19. Slide length 1.5m 30°?

20. Kite string 60m 60°?

Part B: 4 Marks Questions (Answers in ~6 Lines)

1. Explain line of sight with example.

2. Difference elevation vs depression.

3. Information for minar height.

4. Tower 15m away 60° height.

5. Ladder for 3.7m at 60°.

6. Chimney with 1.5m observer 45°.

7. Flag on 10m building 30°/45°.

8. Shadow 40m longer 30°/60°.

9. Multi-building 30°/45° 8m.

10. River 3m bridge 30°/45°.

11. Rope 20m 30° height.

12. Tree broken 30° 8m.

13. Slide 1.5m 30° length.

14. Tower 30m away 30° height.

15. Kite 60m height 60° string.

16. Boy walks 30° to 60° distance.

17. Transmission tower height.

18. Statue on pedestal 60°/45°.

19. Building tower 30°/60° 50m.

20. Poles equal 60°/30° 80m road.

Part C: 8 Marks Questions (Detailed Answers)

1. Explain angle of elevation with diagram.

2. Explain angle of depression with diagram.

3. Process to find minar height.

4. Solve tower 15m 60°.

5. Solve ladder pole repair.

6. Solve chimney observer.

7. Solve flag building.

8. Solve shadow tower.

9. Solve multi-building.

10. Solve river bridge.

11. Solve circus rope.