Some Applications of Trigonometry Class 10 Maths Chapter 9 CBSE Notes - 2025-26

Angles of Elevation and Depression:

• The line of sight is a line drawn from an observer's eye to a point in the thing being observed.

• Angle of Elevation:

When the point being observed is above the horizontal level, the angle created by the line of sight with the horizontal is called the angle of elevation of the point being viewed.

• Let $\text{O}$ and $\text{P}$ be two points, with $\text{P}$ being the greater level. Assume that $\text{OA}$ and $\text{PB}$ are horizontal lines that pass through $\text{O}$ and $\text{P}$, respectively.

• If the observer is at $\text{O}$ and the object is at $\text{P}$, the line $\text{OP}$ is known as the line of sight of Point $\text{P}$, and the angle $\text{AOP}$ between the line of sight and the horizontal line $OA$ is known as the angle of elevation of Point $\text{P}$ as seen from $\text{O}$.

• Angle of Depression:

The angle created by the line of sight with the horizontal when the point is below the horizontal level is known as the angle of depression of a point on the object being viewed.

• The angle $\text{BPO}$ is known as the angle of depression of $\text{O}$ as seen from $\text{P}$ if the observer is at $\text{P}$ and the object under examination is at $\text{O}$.

• From the diagram, it is obviously clear that the angle of elevation of a point $\text{P}$when viewed from a point $\text{O}$ is equal to the angle of depression of $\text{O}$ when viewed from $\text{P}$.

Example:

1. From the top of a $\text{7 m}$ high building, the angle of elevation of the top of a cable tower is \[\text{6}{{\text{0}}^{\text{0}}}\] and the angle of depression of its foot is \[{{45}^{\text{0}}}\] . Determine the height of the tower.

Ans:

Given Data:

Given the height of building $\text{= 7 m}$

Angle of the elevation of top of a tower from the top of building \[\text{= 6}{{\text{0}}^{\text{0}}}\]

Angle of the depression of the foot of a tower \[\text{= 4}{{\text{5}}^{\text{0}}}\]

To find:

We need to calculate the height of the tower, that is \[\text{AE}\]

Sol:

From the given data, we have

Angle of elevation, that is 

\[\angle \text{ACB = 6}{{\text{0}}^{\text{0}}}\]

Angle of depression, that is

\[\angle \text{BCE = 4}{{\text{5}}^{\text{0}}}\]

Here,

$BE = CD = 7 m$

In the right angled triangle \[ABC\] ,

\[\text{tan }\!\!\theta\!\!\text{  = }\dfrac{\text{Opposite Side}}{\text{Adjacent Side}}\]

Therefore,

$\text{tan 6}{{\text{0}}^{0}}\text{ = }\dfrac{\text{Opposite Side}}{\text{Adjacent Side}}$ 

$\text{tan 6}{{\text{0}}^{0}}\text{ = }\dfrac{\text{h}}{\text{CB}}$

Thus,

\[\text{h = CB }\!\!\times\!\!\text{ tan 6}{{\text{0}}^{\text{0}}}\]

Since,

\[\text{tan 6}{{\text{0}}^{\text{0}}}=\sqrt{3}\]

Therefore,

$\text{h = CB }\!\!\times\!\!\text{ }\sqrt{\text{3}}$

$\text{h = }\sqrt{\text{3}}\text{ CB          ------}\left( \text{1} \right)$

In the right angled triangle \[CBE\] ,

\[\text{tan }\!\!\theta\!\!\text{  = }\dfrac{\text{Opposite Side}}{\text{Adjacent Side}}\]

Therefore,

$\text{tan 4}{{\text{5}}^{0}}\text{ = }\dfrac{\text{Opposite Side}}{\text{Adjacent Side}}$

$\text{tan 4}{{\text{5}}^{0}}\text{ = }\dfrac{7}{\text{CB}}$

Since,

\[\text{tan 4}{{\text{5}}^{\text{0}}}=1\]

Therefore,

$\text{   7 = CB }\!\!\times\!\!\text{ 1}$

$\text{CB = 7          ------}\left( \text{2} \right)$

By putting the equation \[\left( 2 \right)\] in the equation \[\left( 1 \right)\], we get

$\text{   h = 7 }\!\!\times\!\!\text{ }\sqrt{\text{3}}$

$\text{AB = 7}\sqrt{\text{3}}$

Therefore, total height of the tower can be calculated as,

Total height of the tower,

$\text{AE = AB + BE}$

$\text{AE = 7}\sqrt{3}+7$

$\text{AE = 7}\left( \sqrt{3}+1 \right)$

Therefore, the total height of the tower is \[\text{7}\left( \sqrt{3}+1 \right)\] unit.

Class 10 Maths Chapter 9 Revision Notes

Class 10 Maths Notes of Some Applications of Trigonometry

Trigonometry is a mathematical concept that we apply on angles and sides of triangles and it serves as a very important and scoring topic for the Class 10 syllabus. The chapters and their weightage for Class 10 board exams are mentioned below in the table:

You will learn how to apply trigonometry concepts on triangles, how to deal with trigonometric ratios, angles and sides, angle of elevation, angle of depression, etc. in this chapter. With the help of various trigonometric formulas and concepts, you will be able to solve statement-based questions of this chapter as well. We will cover Class 10 Maths Chapter 9 revision notes in this article and will see the various benefits of these notes:

Benefits of Revision Notes Class 10 Maths Chapter 9

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• Class 10 Revision Notes Some Applications of Trigonometry will help you in covering the important and scoring topics of the syllabus that is Trigonometry and Chapter 9 is considered as one of the toughest and scoring chapters and these notes will help you in preparing this topic.

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General Tips

• While revising, go through every point and concept related to the chapter carefully in Ch 9 Class 10 Maths revision notes because it is one of the toughest chapters of Class 10.

• Prefer revising it every time before you start solving the problems of the chapters because in these chapters you have to deal with statement-based questions and need to make their diagrams as well, only then you will be able to solve the question at the end.

• Do also revise the notes of Chapter 8 related to Introduction to Trigonometry because some of the basic concepts of that chapter will also be used in Chapter 9.

• You can revise it again in future grades before you start with the Trigonometry chapter in upper classes as well.

Solved Questions

1. The angle of elevation of the top of a tower from a point on the ground, which is at a distance of 30 m from the foot of the tower, is 30°. Determine the height of the tower.

Solution:

Let AB be the height of the tower and C be the point of elevation, which is at a distance of 30 m from the foot of the tower.

(Image Will Be Updated Soon)

In right ⊿ ABC

tan 30° = AB/BC

1/√3 = AB/30

⇒ AB = 10√3

Therefore, the height of the tower is 10√3 m.

2. A kite flies at a height of 60 m above the level of the ground. Its string is temporarily tied to a point on the ground. The string’s inclination with the ground is at 60°. Determine the length of this string, considering that the string doesn’t have any slack.

Solution:

Sketch out a figure based on the given information.

(Image Will Be Updated Soon)

From the figure BC = Height of the kite from the ground = 60 m, AC = Inclined length of the string from the ground, and A be the point where the string of the kite is tied.

From the drawn figure,

sin 60° = BC/AC

⇒ √3/2 = 60/AC

⇒ AC = 40√3 m

Therefore, the length of the string from the ground is 40√3 m.

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Incorporated in the CBSE Class 10 Maths syllabus, this Trigonometry chapter emphasizes a practical approach, aiding students in mastering its application in daily life and exams. To facilitate easy comprehension, our expert teachers have curated study materials that simplify the topic without unnecessary complexity. We recommend students delve into these revision notes and explore related links in this article to optimize their learning experience and make the most of the valuable insights provided by our experts.

Conclusion:

When it comes to Class 10 Maths, Trigonometry stands out as a crucial and high-scoring topic, comprising two vital chapters. Questions from this area are a recurring feature in exams, emphasizing the need for thorough practice. Vedantu addresses this with Class 10 Maths Notes on Some Applications of Trigonometry, aiding in last-minute revision. Tailored to the syllabus and board exam requirements, these notes save time, allowing students to concentrate on studies. Designed for those aiming for good marks, Vedantu's notes ensure a quick recap without missing any details, empowering students to confidently solve problems from this chapter.

Related Study Materials for Class 10 Maths Chapter 9 Some Applications of Trigonometry

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