Class 10 Maths Chapter 9 – Applications of Trigonometry | NCERT Solutions, Angle of Elevation & CBSE Questions (2025–26)

 Class 10 CBSE Maths – Chapter 9: Some Applications of Trigonometry

---

🔹 Key Concepts

✅ Line of Sight

• A line drawn from the eye of an observer to the point being viewed.

✅ Angle of Elevation

• The angle between the horizontal and the line of sight when the point being viewed is above the horizontal level.

✅ Angle of Depression

• The angle between the horizontal and the line of sight when the point being viewed is below the horizontal level.

---

📘 Trigonometric Ratios Used

In all questions involving height and distance, use these ratios:
$\sin\theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}}, \quad \cos\theta = \frac{\text{Base}}{\text{Hypotenuse}}, \quad \tan\theta = \frac{\text{Perpendicular}}{\text{Base}}$

Most problems in this chapter use right-angled triangles and involve tan θ.

---

🧠 Important Notes:

• Always draw a diagram to represent the situation.

• Use trigonometric ratios to form equations.

• Heights, distances, and angles must be consistent with the diagram.

---

📄 Exercise 9.1 – All Questions from NCERT

Q1. A tower stands vertically on the ground. From a point on the ground 20 m away from the foot of the tower, the angle of elevation of the top of the tower is 60°. Find the height of the tower.

Solution:

• Given: Base = 20 m, angle = 60°

• $\tan 60° = \frac{h}{20} \Rightarrow \sqrt{3} = \frac{h}{20} \Rightarrow h = 20\sqrt{3} \approx 34.64 \text{ m}$

---

Q2. A man on a cliff observes a boat at an angle of depression of 30°. If the cliff is 50 m high, find the distance of the boat from the base of the cliff.

Solution:

• Given: Height = 50 m, angle = 30°

• $\tan 30° = \frac{50}{x} \Rightarrow \frac{1}{\sqrt{3}} = \frac{50}{x} \Rightarrow x = 50\sqrt{3} \approx 86.6 \text{ m}$

---

Q3. The angle of elevation of the top of a tower from a point on the ground is 30°. On walking 100 m towards the tower, the angle of elevation becomes 60°. Find the height of the tower.

Solution:
Let height = h and original distance = x

• $\tan 30° = \frac{h}{x} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{x} \Rightarrow h = \frac{x}{\sqrt{3}}$

• After walking 100 m:
  $\tan 60° = \frac{h}{x - 100} \Rightarrow \sqrt{3} = \frac{h}{x - 100} \Rightarrow h = \sqrt{3}(x - 100)$

Equating:
$\frac{x}{\sqrt{3}} = \sqrt{3}(x - 100) \Rightarrow x = 150 \Rightarrow h = \frac{150}{\sqrt{3}} \approx 86.6 \text{ m}$

---

🔁 Summary and Revision Notes

• Use tan θ frequently in problems

• Always label right-angled triangle with appropriate lengths and angles

• Use standard values: $\tan 30° = 1/\sqrt{3}, \tan 45° = 1, \tan 60° = \sqrt{3}$

---

📘 CBSE Previous Year Questions – Applications of Trigonometry

✅ 1-Mark

(2018) Define angle of elevation.
📌 The angle formed by the line of sight with the horizontal when the point is above the level of the observer.

✅ 2-Mark

(2020) From a point 40 m away from the base of a tower, the angle of elevation of its top is 45°. Find the height of the tower.

$$\tan 45° = \frac{h}{40} \Rightarrow h = 40 \text{ m}$$

✅ 3-Mark

(2023) A person 1.5 m tall is standing 30 m away from a building. The angle of elevation of the top of the building is 60°. Find the height of the building.
Let total height be h.

$$\tan 60° = \frac{h - 1.5}{30} \Rightarrow \sqrt{3} = \frac{h - 1.5}{30} \Rightarrow h = 30\sqrt{3} + 1.5 \approx 53.48 \text{ m}$$