CBSE Class 10 Mathematics NCERT Solutions Class 10 Maths Chapter 9 Applications of Trigonometry (2025–2026)

Mastering CBSE Class 10 Mathematics NCERT Solutions Class 10 Maths Chapter 9 Applications of Trigonometry is essential for securing a top score in your upcoming 2025–2026 board exam 10. Also known as "Heights and Distances," this chapter bridges pure trigonometric ratios with real-world geometric measurements, such as calculating the height of towers, the distance of ships from a lighthouse, or the altitude of flying aircraft without physical measurement. This guide provides comprehensive concept breakdowns, standard trigonometric values, solved board-level exemplar problems, and targeted preparation strategies designed strictly according to the latest CBSE Mathematics curriculum.

Key Concepts & Core Formulas

Chapter 9 relies on right-angled triangle geometry and three fundamental concepts: the line of sight, the angle of elevation, and the angle of depression. Understanding these definitions is the first step to drawing correct figures.

1. Line of Sight

The line of sight is the straight line drawn from the eye of an observer to the point in the object viewed by the observer.

2. Angle of Elevation

When an observer looks up at an object located above the horizontal eye level, the angle formed by the line of sight with the horizontal line is called the angle of elevation.

• Condition: Object is positioned higher than the observer's eye level.
• Formula Context: In right ΔABC (where ∠B = 90°, AB is vertical height, and BC is horizontal ground), if θ is the angle of elevation at point C, then tan θ = Opposite / Adjacent = AB / BC.

3. Angle of Depression

When an observer looks down at an object located below the horizontal eye level, the angle formed by the line of sight with the horizontal line is called the angle of depression.

• Critical Geometric Property: The horizontal line drawn at the observer's eye level is parallel to the ground level. Therefore, by the alternate interior angles theorem, the angle of depression of an object from the observer is strictly equal to the angle of elevation of the observer from the object.

4. Essential Trigonometric Ratios Table

Most CBSE Class 10 board exam questions revolve around standard angles: 30°, 45°, and 60°. Memorizing this table ensures swift and error-free problem-solving:

Standard constant values to remember: √3 ≈ 1.732 and √2 ≈ 1.414 (use only when specified in the question).

Step-by-Step Problem-Solving Strategy

1. Read and Diagram: Draw a clean, labeled right-angled triangle representing the scenario. Always mark vertical elements (towers, trees, buildings) at 90° to the horizontal base.
2. Incorporate Observer Height: If the observer's height is given (e.g., 1.5 m), subtract it from the total vertical height to find the opposite side of the working triangle. If no observer height is mentioned, treat the observer as a point on the ground.
3. Select the Right Ratio: If opposite and adjacent sides are involved, use tan θ. If hypotenuse (ladder length, kite string, rope) is involved, use sin θ or cos θ.
4. Set up Simultaneous Equations: For multi-triangle problems, express the common side (usually the horizontal distance or vertical height) in terms of the unknown variable across both triangles.
5. Rationalize Denominators: Never leave radicals in the denominator of your final answer (e.g., write 10/√3 as (10√3)/3 m).

Important CBSE Questions with Step-by-Step Solutions

Question 1 (Standard 2-Mark 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 found to be 60°. Find the height of the tower.

Solution:

1. Let AB be the vertical tower of height h meters.
2. Let C be the point on the ground such that distance BC = 15 m.
3. The angle of elevation ∠ACB = 60°. ΔABC is right-angled at B.
4. In right ΔABC:
   tan 60° = AB / BC
√3 = h / 15
h = 15√3 m
5. If √3 = 1.732, then h = 15 × 1.732 = 25.98 m.

Answer: The height of the tower is 15√3 m (or 25.98 m).

Question 2 (3-Mark Question: Building and Cable Tower)

From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

Solution:

1. Let AB be the building of height 7 m and CD be the cable tower of height H.
2. Let the distance between the bases be BD = x m.
3. Draw a horizontal line AE parallel to BD intersecting CD at E.
   Thus, ED = AB = 7 m and CE = H - 7 m. Also, AE = BD = x m.
4. From the top of the building (point A):
   • Angle of depression to foot D: ∠EAD = 45° ⇒ ∠ADB = 45° (alternate interior angles).
   • Angle of elevation to top C: ∠CAE = 60°.
5. In right ΔABD:
   tan 45° = AB / BD
1 = 7 / x ⇒ x = 7 m
   Therefore, AE = 7 m.
6. In right ΔCEA:
   tan 60° = CE / AE
√3 = (H - 7) / 7
H - 7 = 7√3
H = 7√3 + 7 = 7(√3 + 1) m
7. Substituting √3 = 1.732:
   H = 7(1.732 + 1) = 7(2.732) = 19.124 m.

Answer: The total height of the cable tower is 7(√3 + 1) m (or 19.12 m).

Question 3 (4-Mark Long Answer / Case-Based: Moving Car on Highway)

A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

Solution:

1. Let AB be the tower of height h. Let B be the foot of the tower on the highway.
2. Let the initial position of the car be C and the position after 6 seconds be D.
3. The angles of depression are 30° and 60°. Therefore, the corresponding angles of elevation from ground level are ∠ACB = 30° and ∠ADB = 60°.
4. Let the uniform speed of the car be v m/s.
   Distance CD = Speed × Time = 6v meters.
   Let the time taken to travel from D to B be t seconds.
   Distance DB = v × t = vt meters.
   Total distance CB = CD + DB = 6v + vt = v(6 + t).
5. In right ΔABD:
   tan 60° = AB / DB
√3 = h / (vt) ⇒ h = vt√3 — (Equation 1)
6. In right ΔABC:
   tan 30° = AB / CB
1/√3 = h / [v(6 + t)] ⇒ h = [v(6 + t)] / √3 — (Equation 2)
7. Equating Equation 1 and Equation 2:
   vt√3 = [v(6 + t)] / √3
   Dividing both sides by v (since v ≠ 0):
   t√3 × √3 = 6 + t
3t = 6 + t
2t = 6 ⇒ t = 3 \text{ seconds}

Answer: The time taken by the car to reach the foot of the tower from point D is 3 seconds.

Question 4 (Important Proof-Based Question)

The angles of elevation of the top of a tower from two points at a distance of a and b from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is √(ab).

Solution:

1. Let AB be the vertical tower of height h.
2. Let C and D be two points on the ground such that BC = a and BD = b (assume a < b).
3. Since the angles of elevation are complementary, if ∠ACB = θ, then ∠ADB = 90° - θ.
4. In right ΔABC:
   tan θ = AB / BC = h / a — (Equation 1)
5. In right ΔABD:
   tan (90° - θ) = AB / BD = h / b
   Since tan (90° - θ) = cot θ:
   cot θ = h / b — (Equation 2)
6. Multiply Equation 1 and Equation 2:
   tan θ × cot θ = (h / a) × (h / b)
   Since tan θ × cot θ = 1:
   1 = h² / (ab)
h² = ab ⇒ h = √(ab) (height cannot be negative).

Conclusion: Hence proved, the height of the tower is √(ab).

Common Mistakes to Avoid in CBSE Board Exams

• Confusing Elevation and Depression Angles: Students often mistakenly mark the angle of depression between the vertical line and the line of sight. Remember: angles are always measured with respect to the horizontal line.
• Ignoring Observer Height: If a problem mentions "a 1.5 m tall observer," do not forget to add this height to the computed triangle side to find the total height of the object.
• Selecting Inefficient Trigonometric Ratios: In 90% of heights and distances problems, tan θ is the ideal ratio because problems involve base distance and vertical height. Only choose sin θ or cos θ when hypotenuse lengths (ropes, ladders, kites) are involved.
• Omitting Units: Forgetting to write meters (m) or centimeters (cm) in intermediate steps and final statements results in unnecessary half-mark deductions.

How to Prepare for This Topic for 2025–2026 Board Exams

1. Master Diagram Drawing: Spend 15 minutes daily sketching figures purely from word descriptions without solving the arithmetic. If your diagram is correct, 70% of the problem is solved.
2. Practice Multi-Triangle Systems: Focus on questions involving two distinct elevation angles (e.g., 30° to 60° or 30° to 45°) and situations with combined elevations and depressions.
3. Solve Previous Years' Question Papers (PYQs): Chapter 9 regularly carries 4 to 5 marks in Section D or Case Study-based questions in the Class 10 CBSE Mathematics paper.

Where to Practice More

To test your speed, accuracy, and diagram-drawing skills under timed exam conditions, generate customized practice worksheets and full-length CBSE Class 10 mock test papers on Theorify QPTool. Access real-time syllabus-aligned chapter assessments, detailed marking schemes, and standard question banks for 2025–2026 board exam preparation.