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# NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 2024-25

Last updated date: 11th Sep 2024
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## NCERT Solutions for Maths Chapter 9 Some Applications of Trigonometry Class 10

The NCERT Solutions for Class 10 Maths Chapter 9, "Some Applications of Trigonometry," offers comprehensive solutions for every question presented in the NCERT Textbook. Chapter 9 consists of a discussion of basic trigonometry, heights, and distance, applications of trigonometry, Trigonometry Ratios, Angle of Elevation, and  Angle of Depression. Prepared and reviewed by subject matter experts, these solutions are regularly updated to align with the latest CBSE Syllabus for the academic year 2024-25 and adhere to the guidelines set by the CBSE board.

Table of Content
1. NCERT Solutions for Maths Chapter 9 Some Applications of Trigonometry Class 10
2. Glance of NCERT Solutions for Chapter 9 Maths Some Applications of Trigonometry Class 10
3. Access Exercise Wise NCERT Solutions for Chapter 9 Maths Class 10
4. Exercises under NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry
5. Access NCERT Solutions for Class 10 Maths Chapter 9 - Some Applications Of Trigonometry
6. Overview of the Exercises Covered in the NCERT Solutions Class 10 Maths Chapter- 9
6.1What is Trigonometry?
6.2Trigonometry Ratios
6.3Why are Some Applications of Trigonometry Important?
7. Important Terms to Remember in Height and Distance
8. Why are Some Applications of Trigonometry Important?
9. Overview of Deleted Syllabus for CBSE Chapter 9 Maths Class 10
10. Class 10 Maths Chapter 9: Exercises Breakdown
11. Other Study Material for CBSE Class 10 Maths Chapter 9
12. Chapter-Specific NCERT Solutions for Class 10 Maths
FAQs

## Glance of NCERT Solutions for Chapter 9 Maths Some Applications of Trigonometry Class 10

• Chapter 9 of Class 10 Maths, "Some Applications of Trigonometry," deals with using trigonometric ratios (sine, cosine, tangent) to solve real-world problems involving heights and distances.

• Angle of elevation and depression: These are angles formed between the horizontal line and the line of sight to an object, above or below the horizontal respectively.

• Usage of the basic definitions of trigonometric ratios sine (sin), cosine (cos), and tangent (tan) in a right triangle ABC (where angle C is 90 degrees):

• sin A = Opposite over Hypotenuse = BC/AB

• cos A = Adjacent over Hypotenuse = AC/AB

• tan A = Opposite over Adjacent = BC/AC

• This article contains chapter notes, formulas, exercise links, and important questions for chapter 9 - Some Applications of Trigonometry.

## Access Exercise Wise NCERT Solutions for Chapter 9 Maths Class 10

 Current Syllabus Exercises of Class 10 Maths Chapter 9 NCERT Solutions of Class 10 Maths Some Applications of Trigonometry Exercise 9.1
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## Exercises under NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry

Exercise 9.1: This exercise covers four questions related to finding angles of elevation and depression. The questions involve finding the height, distance, or both of an object using trigonometric ratios. The solutions provided aim to help students understand the practical application of trigonometry in real-life situations.

## Access NCERT Solutions for Class 10 Maths Chapter 9 - Some Applications Of Trigonometry

Exercise- 9.1

1. A circus artist is climbing a $20m$ long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is $\mathbf{30}{}^\circ$.

Ans: By observing the figure, $AB$ is the pole.

In$\Delta ABC$,

$\frac{\text{AB}}{\text{AC}}=\sin {{30}^{{}^\circ }}$

$\Rightarrow \frac{\text{AB}}{20}=\frac{1}{2}$

$\Rightarrow \text{AB}=\frac{20}{2}=10$

Therefore, the height of the pole is$10m$.

2. A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle $\mathbf{30}{}^\circ$with it. The distance between the foot of the tree to the point where the top touches the ground is $\mathbf{8m}$. Find the height of the tree.

Ans: Let $AC$was the original tree. Due to the storm, it was broken into two parts. The broken part  $AB$ is making 30° with the ground.

Let $\mathrm{AC}$ be the original tree. Due to the storm, it was broken into two parts. The broken part $\mathrm{A}^{\prime} \mathrm{B}$ is making ${{30}^{{}^\circ }}$ with the ground. In triangle${{\text{A}}^{\prime }}\text{BC}$,

$\Rightarrow \frac{BC}{{{A}^{\prime }}C}=\tan {{30}^{{}^\circ }}$

$\Rightarrow \frac{BC}{8}=\frac{1}{\sqrt{3}}$

$\Rightarrow \text{BC}=\left( \frac{8}{\sqrt{3}} \right)\text{m}$

$\Rightarrow \frac{{{\text{A}}^{\prime }}\text{C}}{{{\text{A}}^{\prime }}\text{B}}=\cos 30$

$\Rightarrow \frac{8}{{{A}^{\prime }}B}=\frac{\sqrt{3}}{2}$

$\Rightarrow {{A}^{\prime }}B=\left( \frac{16}{\sqrt{3}} \right)m$

Height of tree $=\mathrm{A}^{\prime} \mathrm{B}+\mathrm{BC}$

$=\left(\frac{16}{\sqrt{3}}+\frac{8}{\sqrt{3}}\right) \mathrm{m}=\frac{24}{\sqrt{3}} \mathrm{~m}$

$=8 \sqrt{3} \mathrm{~m}$

3. A contractor plans to install two slides for the children to play in a park. For the children below the age of $5$ years, she prefers to have a slide whose top is at a height of$\mathbf{1}.\mathbf{5m}$, and is inclined at an angle of $\mathbf{30}{}^\circ$ to the ground, whereas for the elder children she wants to have a steep slide at a height of$\mathbf{3m}$, and inclined at an angle of $60{}^\circ$ to the ground. What should be the length of the slide in each case?

Ans: It can be observed that $\text{AC}$ and $\text{PR}$ are the slides for younger and elder children respectively.

In $\vartriangle \text{ABC}$

$\Rightarrow \frac{\text{AB}}{\text{AC}}=\sin 30$

$\Rightarrow \frac{1.5}{\text{AC}}=\frac{1}{2}$

$\Rightarrow \text{AC}=3~\text{m}$

In $\vartriangle \text{PQR}$,

$\Rightarrow \frac{\text{PQ}}{\text{PR}}=\sin {{60}^{{}^\circ }}$

$\Rightarrow \frac{3}{\text{PR}}=\frac{\sqrt{3}}{2}$

$\Rightarrow \text{PR}=\frac{6}{\sqrt{3}}=2\sqrt{3}~\text{m}$

Therefore, the lengths of these slides are $3~\text{m}$ and $2\sqrt{3}~\text{m}$.

4. The angle of elevation of the top of a tower from a point on the ground, which is$\mathbf{30m}$ away from the foot of the tower is $\mathbf{30}{}^\circ .$ Find the height of the tower.

Ans: Let $\mathrm{AB}$ be the tower and the angle of elevation from point $\mathrm{C}$ (on ground) is $30^{\circ}$

In $\vartriangle \text{ABC}$ ,

$\Rightarrow \frac{\text{AB}}{\text{BC}}=\tan {{30}^{{}^\circ }}$

$\Rightarrow \frac{\text{AB}}{30}=\frac{1}{\sqrt{3}}$

$\Rightarrow \text{AB}=\frac{30}{\sqrt{3}}=10\sqrt{3}~\text{m}$

Therefore, the height of the tower is $10\sqrt{3}~\text{m}$.

5. A kite is flying at a height of $\mathbf{60m}$ above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is $\mathbf{60}{}^\circ .$ Find the length of the string, assuming that there is no slack in the string.

Ans: Let $\text{K}$ be the kite and the string is tied to point $\text{P}$ on the ground.

In $\vartriangle \text{KLP}$,

$\Rightarrow \frac{\text{KL}}{\text{KP}}=\sin {{60}^{{}^\circ }}$

$\Rightarrow \frac{60}{\text{KP}}=\frac{\sqrt{3}}{2}$

$\Rightarrow \text{KP}=\frac{120}{\sqrt{3}}=40\sqrt{3}~\text{m}$

Hence, the length of the string is $40\sqrt{3}~\text{m}$.

6. A $\mathbf{1}.\mathbf{5m}$ tall boy is standing at some distance from a $\mathbf{30m}$ tall building. The angle of elevation from his eyes to the top of the building increases from $\mathbf{30}{}^\circ$ to $\mathbf{60}{}^\circ$ as he walks towards the building. Find the distance he walked towards the building.

Ans : Let the boy was standing at point S initially. He walked towards the building and reached at point T.

$\text{PR}=\text{PQ}-\text{RQ}$

$=(30-1.5)=28.5~=\frac{57}{2}~$

In $\vartriangle \text{PAR}$,

$\Rightarrow \frac{\text{PR}}{\text{AR}}=\tan {{30}^{{}^\circ }}$

$\Rightarrow \frac{57}{2\text{AR}}=\frac{1}{\sqrt{3}}$

$\Rightarrow \text{AR}=\left( \frac{57}{2}\sqrt{3} \right)\text{m}$

In $\vartriangle \text{PRB}$,

$\Rightarrow \frac{\text{PR}}{\text{BR}}=\tan {{60}^{{}^\circ }}$

$\Rightarrow \frac{57}{2B\text{R}}=\sqrt{3}$

$\Rightarrow \text{BR}=\frac{57}{2\sqrt{3}}\text{=}\frac{19\sqrt{3}}{2}m$

By observing the figure,

$ST=AB$

$=AR-BR=\left( \frac{57\sqrt{3}}{2}-\frac{19\sqrt{3}}{2} \right)$

$=\left( \frac{38\sqrt{3}}{2} \right)=19\sqrt{3}m$

Hence, he walked $19\sqrt{3}m$ towards the building.

7. From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a $\mathbf{20m}$ high building are $\mathbf{45}{}^\circ$ and $\mathbf{60}{}^\circ$ respectively. Find the height of the tower.

Ans: Let $AB$ be the statue, $BC$ be the pedestal, and $D$ be the point on the ground from where the elevation angles are to be measured.

In $\Delta \text{BCD}$,

$\Rightarrow \frac{\text{BC}}{\text{CD}}=\tan {{45}^{{}^\circ }}$

$\Rightarrow \frac{\text{BC}}{\text{CD}}=1$

$\Rightarrow \text{BC}=\text{CD}$

In $\Delta A\text{CD}$,

$\Rightarrow \frac{\text{AB}+\text{BC}}{\text{CD}}=\tan {{60}^{{}^\circ }}$

$\Rightarrow \frac{\text{AB}+\text{BC}}{\text{CD}}=\sqrt{3}$

$\frac{AB+20}{20}=\sqrt{3}$

$AB=\left( 20\sqrt{3}-20 \right)m$

$=20\left( \sqrt{3}-1 \right)m$

8.  A statue, $\mathbf{1}.\mathbf{6m}$ tall, stands on a top of pedestal, from a point on the ground, the angle of  elevation of the top of statue is $\mathbf{60}{}^\circ$ and from the same point the angle of elevation of the  top of the pedestal is $\mathbf{45}{}^\circ .$ Find the height of the pedestal.

Ans:

Let AB be the statue, BC be the pedestal, and D be the point on the ground from where  the elevation angles are to be measured.

$\text{ In }\vartriangle \text{BCD,}$

$\text{ }\frac{\text{BC}}{\text{CD}}=\tan 45$

$\frac{\text{BC}}{\text{CD}}=1$

$\text{BC}=\text{CD }$

$\text{In }\vartriangle \text{ACD, }$

$\frac{\text{AB}+\text{BC}}{\text{CD}}=\tan {{60}^{{}^\circ }}$

$\frac{\text{AB}+\text{BC}}{\text{CD}}=\sqrt{3}\text{ }$

$1.6+\text{BC}=\text{BC}\sqrt{3}\quad [\text{As}\,\,\text{CD}=\text{BC}]\,\,$

$\text{BC}(\sqrt{3}-1)=1.6\,\,\,$

$\text{BC}=\frac{(1.6)(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}\quad [\text{ByRationalization}]$

$=\frac{1.6(\sqrt{3}+1)}{{{(\sqrt{3})}^{2}}-{{(1)}^{2}}}$

$=\frac{1.6\left( \sqrt{3}+1 \right)}{2}=0.8\left( \sqrt{3}+1 \right)m$

9. The angle of elevation of the top of a building from the foot of the tower is $\mathbf{30}{}^\circ$ and the angle of elevation of the top of the tower from the foot of the building is $\mathbf{60}{}^\circ .$ If the tower is $\mathbf{50m}$ high, find the height of the building.

Ans: Let $AB$be the building and $CD$be the tower.

In $\Delta \text{CDB}$,

$\Rightarrow \frac{\text{CD}}{\text{BD}}=\tan {{60}^{{}^\circ }}$

$\Rightarrow \frac{50}{\text{BD}}=\sqrt{3}$

$\Rightarrow \text{BD}=\frac{50}{\sqrt{3}}$

In $\Delta ABD$

$\Rightarrow \frac{AB}{BD}=\tan 30{}^\circ$

$\Rightarrow AB=\frac{50}{\sqrt{3}}\left( \frac{1}{\sqrt{3}} \right)=\frac{50}{3}=16\frac{2}{3}$

Therefore, the height of the building is $16\frac{2}{3}$m.

10. Two poles of equal heights are standing opposite each other on either side of the road, which is $\mathbf{80m}$wide. From a point between them on the road, the angles of elevation of the top of the poles are $\mathbf{60}{}^\circ$and $\mathbf{30}{}^\circ$respectively. Find the height of poles and the distance of the point from the poles.

Ans: Let $AB$ and $CD$ be the poles and $O$ is the point from where the elevation angles are measured.

In $\Delta \mathrm{CDO}$,

$\Rightarrow \frac{\text{AB}}{\text{BO}}=\tan {{60}^{{}^\circ }}$

$\Rightarrow \frac{\text{AB}}{\text{BO}}=\sqrt{3}$

$\Rightarrow \text{BO}=\frac{\text{AB}}{\sqrt{3}}$

In $\Delta \text{CDO}$,

$\Rightarrow \frac{\text{CD}}{\text{DO}}=\tan {{30}^{{}^\circ }}$

$\Rightarrow \frac{\text{CD}}{80-\text{BO}}=\frac{1}{\sqrt{3}}$

$\Rightarrow \text{CD}\sqrt{3}=80-\text{BO}$

$\Rightarrow \text{CD}\sqrt{3}=80-\frac{\text{AB}}{\sqrt{3}}$

$\Rightarrow \text{CD}\sqrt{3}+\frac{\text{AB}}{\sqrt{3}}=80$

Since the poles are of equal heights, $\text{CD}=\text{AB}$

$\Rightarrow \text{CD}\left[ \sqrt{3}+\frac{1}{\sqrt{3}} \right]=80$

$\Rightarrow \text{CD}\left( \frac{3+1}{\sqrt{3}} \right)=80$

$\Rightarrow \text{CD}=20\sqrt{3}~\text{m}$

By observing the figure,

$\Rightarrow BO=\frac{AB}{\sqrt{3}}=\frac{CD}{\sqrt{3}}=\frac{20\sqrt{3}}{\sqrt{3}}=20m$

$\Rightarrow DO=BD-BO=80-20=60m$

Therefore, the height of poles is $20\sqrt{3}$ and the point is $20m$ and $60m$ far from these poles.

11. A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower the angle of elevation of the top of the tower is $\mathbf{60}{}^\circ .$ From another point $\mathbf{20m}$away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is $\mathbf{30}{}^\circ .$ Find the height of the tower and the width of the canal.

Ans: In $\Delta \text{ABC}$,

$\Rightarrow \dfrac{\text{AB}}{\text{BC}}=\tan {{60}^{{}^\circ }}$

$\Rightarrow \dfrac{\text{AB}}{\text{BO}}=\sqrt{3}$

$\Rightarrow \text{BC}=\frac{\text{AB}}{\sqrt{3}}$…. (1)

In $\Delta ABD$,

$\Rightarrow \dfrac{AB}{\text{BD}}=\tan {{30}^{{}^\circ }}$

$\Rightarrow \dfrac{AB}{BC+CD}=\frac{1}{\sqrt{3}}$

$\Rightarrow \dfrac{AB}{\frac{AB}{\sqrt{3}}+20}=\dfrac{1}{\sqrt{3}}$

$\Rightarrow \dfrac{AB\sqrt{3}}{AB+20\sqrt{3}}=\dfrac{1}{\sqrt{3}}$

$\Rightarrow 3AB=AB+20\sqrt{3}$

$\Rightarrow 2AB=20\sqrt{3}$

$\Rightarrow AB=10\sqrt{3}m$

Substitute $AB=10\sqrt{3}m$ in $\text{BC}=\dfrac{\text{AB}}{\sqrt{3}}$,

$\Rightarrow BC=\dfrac{AB}{\sqrt{3}}=\dfrac{10\sqrt{3}}{\sqrt{3}}=10m$

Therefore, the height of the tower is $10\sqrt{3}m$ and the width of the canal is $10m$.

12. From the top of a $\mathbf{7m}$ high building, the angle of elevation of the top of a cable tower is $\mathbf{60}{}^\circ$ and the angle of depression of its foot is $\mathbf{45}{}^\circ .$ Determine the height of the tower.

Ans: Let $AB$ be a building and $CD$ be a cable tower.

In $\Delta \text{ABD}$,

$\Rightarrow \frac{\text{AB}}{\text{BD}}=\tan {{45}^{{}^\circ }}$

$\Rightarrow \frac{7}{\text{BD}}=1$

$\Rightarrow \text{BD=7m}$

In $\Delta \text{ACE}$, $AE=BD=7m$

$\Rightarrow \frac{\text{CE}}{AE}=\tan {{60}^{{}^\circ }}$

$\Rightarrow \frac{\text{CE}}{7}=\sqrt{3}$

$\Rightarrow \text{CE=7}\sqrt{3}\text{m}$

$\Rightarrow CD=CE+ED=\left( 7\sqrt{3}+7 \right)=7\left( \sqrt{3}+1 \right)m$

Therefore, the height of the cable tower is $7\left( \sqrt{3}+1 \right)m$.

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

Ans: Let $AB$be the lighthouse and the two ships be at point $C$ and $D$respectively.

In $\Delta \text{ABC}$,

$\Rightarrow \frac{\text{AB}}{\text{BC}}=\tan {{45}^{{}^\circ }}$

$\Rightarrow \frac{75}{\text{BC}}=1$

$\Rightarrow \text{BC=75m}$

In $\Delta \text{ABD}$,

$\Rightarrow \frac{AB}{BD}=\tan {{30}^{{}^\circ }}$

$\Rightarrow \frac{75}{BC+CD}=\frac{1}{\sqrt{3}}$

$\Rightarrow \frac{75}{75+CD}=\frac{1}{\sqrt{3}}$

$\Rightarrow 75\sqrt{3}=75+CD$

$\Rightarrow 75\left( \sqrt{3}-1 \right)m=CD$

Therefore, the distance between the two ships is $75\left( \sqrt{3}-1 \right)m.$

14. A $\mathbf{1}.\mathbf{2m}$ tall girl spots a balloon moving with the wind in a horizontal line at a height of $\mathbf{88}.\mathbf{2m}$ from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is $\mathbf{60}{}^\circ .$ After some time, the angle of elevation reduces to $\mathbf{30}{}^\circ .$ Find the distance travelled by the balloon during the interval.

Ans: Let the initial position $A$ of balloon change to $B$ after some time and $CD$ be the girl.

In $\Delta \text{ACE}$,

$\Rightarrow \frac{\text{AE}}{\text{CE}}=\tan {{60}^{{}^\circ }}$

$\Rightarrow \frac{\text{AF-EF}}{\text{CE}}=\tan {{60}^{{}^\circ }}$

$\Rightarrow \frac{\text{88}\text{.2-1}\text{.2}}{\text{CE}}=\sqrt{3}$

$\Rightarrow \frac{\text{87}}{\text{CE}}=\sqrt{3}$

$\Rightarrow CE=\frac{87}{\sqrt{3}}=29\sqrt{3}m$

In $\Delta B\text{CG}$,

$\Rightarrow \frac{\text{BG}}{\text{CG}}=\tan {{30}^{{}^\circ }}$

$\Rightarrow \frac{\text{BH-GH}}{\text{CG}}=\frac{1}{\sqrt{3}}$

$\Rightarrow \frac{\text{88}\text{.2-1}\text{.2}}{\text{CG}}=\frac{1}{\sqrt{3}}$

$\Rightarrow \text{87}\sqrt{3}=CG$

Distance travelled by balloon$=EG=CG-CE$

$=\left( 87\sqrt{3}-29\sqrt{3} \right)$

$=58\sqrt{3}m$

15. 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 $\mathbf{30}{}^\circ ,$ 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 $\mathbf{60}{}^\circ .$ Find the time taken by the car to reach the foot of the tower from this point.

Ans: Let $AB$ be the tower. Initial position of the car is $C$, which changes to $D$ after six seconds.

In $\Delta \text{ADB}$,

$\Rightarrow \frac{\text{AB}}{CB}=\tan {{60}^{{}^\circ }}$

$\Rightarrow \frac{\text{AB}}{DB}=\sqrt{3}$

$\Rightarrow DB=\frac{AB}{\sqrt{3}}$

In $\Delta \text{ABC}$,

$\Rightarrow \frac{\text{AB}}{BC}=\tan {{30}^{{}^\circ }}$

$\Rightarrow \frac{\text{AB}}{BD+DC}=\frac{1}{\sqrt{3}}$

$\Rightarrow AB\sqrt{3}=BD+DC$

$\Rightarrow AB\sqrt{3}=\frac{AB}{\sqrt{3}}+DC$

$\Rightarrow DC=AB\sqrt{3}-\frac{AB}{\sqrt{3}}=AB\left( \sqrt{3}-\frac{1}{\sqrt{3}} \right)=\frac{2AB}{\sqrt{3}}$

Time taken by the car to travel distance DC $\left( i.e\frac{2AB}{\sqrt{3}} \right)=6$ seconds.

Time taken by the car to travel distance DB $\left( ie.,\frac{AB}{\sqrt{3}} \right)=\frac{6}{\frac{2AB}{\sqrt{3}}}\left( \frac{AB}{\sqrt{3}} \right)=\frac{6}{2}=3$ seconds.

## Overview of the Exercises Covered in the NCERT Solutions Class 10 Maths Chapter- 9

### What is Trigonometry?

Trigonometry is one of the most historical subjects studied by different scholars throughout the world. As you have read in Chapter 8 that trigonometry was introduced because its requirement arose to astronomy. Since then trigonometry is used to calculate the distance from the Earth to the stars and the planets. The most important use of trigonometry is to find out the height of the highest mountain in the world i.e. Mount Everest which is named after Sir George Everest. It is also widely used in Geography and navigation. The knowledge of trigonometry enables us to construct maps, evaluate the position of an island concerning the longitudes and latitudes.

### Historical Facts. Let us Turn to the History of Trigonometry

The trigonometry was used by surveyors for centuries. One of the notable and the largest surveying projects of the nineteenth century was the “ Great Trigonometric Survey" of British India for which the two largest theodolites were constructed. The highest mountain in the world was discovered during the survey in 1852. From a distance of over 160 km, this peak was seen from 6 distinct stations. This peak was named after Sir George Everest who had first used the theodolites. Theodolites are now exhibited in the museum of the surveys of Dehradun.

### Height and Distance

In this topic, you will study about the line of sight, angle of elevation, horizontal level, and angle of depression. All these terms are explained in a detailed form along with some solved examples based on it. These solved examples based on the terms line of sight, angle of elevation and angle of depression will help you to understand the concepts thoroughly.

### How to Calculate Height and Distance?

Trigonometric ratios are used to find out the height and the distance of the object. For example: In figure 1, you can see a boy looking at the top of the lampost. AB is considered as the horizontal level. This level is stated as the line parallel to the ground passing through the viewer's eyes. AC is considered as the line of sight. ∠A is known as the angle of elevation. Similarly, in figure 2, you can see PQ is the line of sight, PR is the horizontal level and ∠P is known as the angle of elevation.

An inclinometer or Clinometer is a device usually used for measuring the angle of elevation and the angle of depression.

Let us recall some trigonometric ratios which help to solve the questions based on  class 10 maths Chapter 9.

### Trigonometry Ratios

The ratio of the sides of a right-angle triangle in terms of any of its acute angle triangle is known as the trigonometric ratio of that specific angle.

In terms of ∠C, the ratio of trigonometry are given as:

Sine - The sine of an angle is stated as the ratio of the opposite side ( perpendicular side) to that angle to the hypotenuse side.

Hence, Sine C = Opposite side/Hypotenuse side

Cosine- The cosine of an angle is stated as the ratio of the adjacent side to that angle to the hypotenuse side.

Hence, Cosine C = Adjacent side /Hypotenuse side

Tangent - The tan of an angle is stated as the ratio of the opposite side (perpendicular side) to that angle to the side adjacent to that angle.

Hence, Tan C = Opposite side/Adjacent side

Cosecant- It is the reciprocal of sine.

Hence, Cosec C = Hypotenuse side/Opposite side

Secant- It is the reciprocal of cosine.

Hence, Sec C= Hypotenuse side/Adjacent side

Cotangent- It is the reciprocal of tangent.

Hence, Cot C = Adjacent side/Opposite side

The following trigonometry ratio table is used to calculate the questions based on applications of trigonometry class 10 NCERT solutions.

Trigonometric Ratio Table

 ∠C 0° 35° 45° 60° 90° Sin C 0 1/2 1/2 3/2 1 Cos C 1 3/2 1/2 1/2 0 Tan C 0 1/3 1 3 Not defined Cosec C Not defined 2 2 2/3 1 Sec C 1 2 / 3 2 2 Not defined Cot C Not defined 3 1 1/3 0

### Why are Some Applications of Trigonometry Important?

Class 10 Chapter 9 some application of trigonometry is an important topic to discuss as it tells how trigonometry is used to find the height and distance of different objects such as the height of the building, the distance between the Earth and Planet and Stars, the height of the highest mountain Mount Everest, etc.

To solve the questions based on some applications of trigonometry class 10, it is necessary to remember trigonometry formulas, trigonometric relations, and values of some trigonometric angles. The following are the concepts covered in the 'height and distance' Some applications of trigonometry.

• To measure the height of big towers or big mountains

• To determine the distance of the shore from the sea.

• To find out the distance between two celestial bodies.

This chapter has a weightage of 12 marks in class 10 Maths Cbse (board) exams.

One question can be expected from this chapter. Class 10 Maths CBSE paper is divided into 4 parts and each question comes with different marks. The questions will be allocated with 1 mark, 2 marks, 3 marks or 4 marks.

Discussion about the sections, exercise, and type of questions given in the exercise.

## Important Terms to Remember in Height and Distance

Line of Sight - It is a line that is drawn from the eye of an observer to the point on the object viewed by the observer.

The Angle of Elevation - It is defined as an angle that is formed between the horizontal line and line of sight. If the line of sight lies upward from the horizontal line, then the angle formed will be termed as an angle of elevation.

Let us take another situation when a boy is standing on the ground and he is looking at the object from the top of the building. The line joining the eye of the man with the top of the building is known as the line of sight and the angle drawn by the line of sight with the horizontal line is known as angle of elevation.

The Angle of Depression - It is defined as an angle drawn between the horizontal line and line of sight. If the line of sight lies downward from the horizontal line, then the angle formed will be termed as an angle of depression.

Let us take a situation when a boy is standing at some height concerning the object he is looking at. In this case, the line joining the eye of the man with the bottom of the building is known as the line of sight and the angle drawn by the line of sight with the horizontal line is known as angle of depression.

Note: Angle of elevation is always equal to the angle of depression

### The Important Point to Remember

The distance of the object is also considered as the base of the right angle triangle drawn through the height of the object and the line of sight.

The length of the horizontal level is also known as the distance of the object it forms the base of the triangle.

Line of sight is considered as a hypotenuse of the right-angle triangle. Hypotenuse side is calculated using Pythagorean Theorem if the height and distance of the object are given.

## Why are Some Applications of Trigonometry Important?

Class 10 Chapter 9 some application of trigonometry is an important topic to discuss as it tells how trigonometry is used to find the height and distance of different objects such as the height of the building, the distance between the Earth and Planet and Stars, the height of the highest mountain Mount Everest, etc.

To solve the questions based on some applications of trigonometry class 10, it is necessary to remember trigonometry formulas, trigonometric relations, and values of some trigonometric angles. The following are the concepts covered in the 'height and distance' Some applications of trigonometry.

• To measure the height of big towers or big mountains

• To determine the distance of the shore from the sea.

• To find out the distance between two celestial bodies.

This chapter has a weightage of 12 marks in class 10 Maths Cbse (board) exams.

One question can be expected from this chapter. Class 10 Maths CBSE paper is divided into 4 parts and each question comes with different marks. The questions will be allocated with 1 mark, 2 marks, 3 marks or 4 marks.

Discussion about the sections, exercise, and type of questions given in the exercise.

The important topic “ Height and Distance" covered in Some applications of trigonometry class 10 is followed by one exercise with 16 questions. The exercise aims to test your knowledge and how deeply you understood each formula and concept of the topic. The numerical questions given in this chapter are based on some applications of trigonometry.

To make you understand the topic and related concept, solved numerical problems are also given. Stepwise solutions are given for each of the solved examples. It will help you to understand which concept and formula will be used to solve the given questions accurately.

## Overview of Deleted Syllabus for CBSE Chapter 9 Maths Class 10

 Chapter Dropped Topics Some Applications of Trigonometry 9.1 Introduction

## Class 10 Maths Chapter 9: Exercises Breakdown

 Exercise Number of Questions Exercise 9.1 Exercise 9.1(15 Questions & Solutions)

## Conclusion

Chapter 9 of Class 10 Maths, "Some Applications of Trigonometry," delves into practical applications of trigonometric concepts in real-life scenarios. Through this chapter, students explore various applications such as heights and distances, navigation, and more. By mastering these applications, students not only enhance their understanding of trigonometry but also recognize its significance in solving everyday problems. Therefore, practicing a variety of problems from NCERT Solutions and previous year papers can enhance preparation and confidence for exams. By mastering Introduction to Trigonometry, students not only excel in mathematics but also build skills that are useful in various real-world scenarios.

## Other Study Material for CBSE Class 10 Maths Chapter 9

 S.No. Important Links for Chapter 9 Some Applications of Trigonometry 1 Class 10 Some Applications of Trigonometry Important Questions 2 Class 10 Some Applications of Trigonometry Revision Notes 3 Class 10 Some Applications of Trigonometry RD Sharma Solutions

## Chapter-Specific NCERT Solutions for Class 10 Maths

Given below are the chapter-wise NCERT Solutions for Class 10 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

NCERT Solutions Class 10 Chapter-wise Maths PDF

## FAQs on NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 2024-25

Vedantu executed well-designed and reviewed applications of trigonometry class 10 ncert solutions which include the latest guidelines as recommended by CBSE board. The solutions are easily accessible from the Vedantiu online learning portal. It can be easily navigated as Vedantu retains a user-friendly network. The detailed solution of all the questions asked in the NCERT book will help you to understand the basic concepts of some applications of trigonometry class 10 thoroughly. Along with the NCERT solutions, the experts of Vedantu also provide various tips and techniques to solve the questions in the exam. Sign in to Vedantu and get access to the ncert solutions for class 10 maths chapter 9 some applications of trigonometry -Free pdf and you can download the updated solution of all the exercises given in the chapter.

2. What is the importance of trigonometry applications in real-life?

The application of trigonometry may not be directly used in solving practical issues but used in distinct fields. For example, trigonometry is used in developing computer music as you must be aware of the fact that sound travels in the form of waves and this wave pattern using sine and cosine functions helps to develop computer music. The following are some of the applications where the concepts of trigonometry and its functions are applicable.

It is used to measure the height and distance of a building or a mountain

It is used in the Aviation sector

It is used in criminology

It is used in creating maps

It is used in satellite systems

The basic trigonometric functions such as sine and cosine are used to determine the sound and light waves.

It is used in oceanography to formulate the height of waves and tides in the ocean.

3. The distance from where the building can be viewed is 90ft from its base and the angle of elevation to the top of the building is 35°. Calculate the height of the building.

Given: Distance from where the building can be viewed is 90ft from its base and angle of elevation to the top of the building is stated as 35°

To calculate the height of the building, we will use the following trigonometry formula

Tan 35° = Opposite Side/ Adjacent Side

Tan 35° = H/90

H = 90 x Tan 35°

H = 90 x 0.7002

H = 63.018 feet

Hence, the height of the building is 63.018 feet.

4. From a 60 meter high tower, the angle of depression of the top and bottom of a house are ‘a and b’ respectively. Calculate the value of x, if the height of the house is [60 sin (β − α)]/ x.

H = d tan β and H- h = d tan α

60/60-h = tan β – tan α

h = [60 tan α – 60 tan β] / [tan β]

h= [60 sin (β – α)/ [(cos α cos β)] [(sin α sin β)]

x = cos α cos β

5. The top of the two different towers of height x and y, standing on the ground level subtended angle of 30° and 60° respectively at the center of the line joining their feet. Calculate x: y.

In ΔABE,

x/a = Tan 30°

x/a = 1/3

x  = a/3

In ΔCDE

y/a = Tan 60°

y/a = 3 → y = a x 3

x/a ÷ y/a =  a/3÷ a3

x/a × y/a =  a/3 × 1/a×3

x/y = 1/3

6. What are the main points to study in Class 10 Maths NCERT Chapter 9?

The main points to study in Class 10 Maths NCERT Chapter 9 include:

• Fundamental Basics Of Trigonometry Applications

• The History Behind Trigonometry

• Concepts On Height And Distance

• The Study About The Line Of Sight

• Angle Of Elevation, Horizontal Level

• Angle Of Depression

• The Calculation Methods For Heights And Distance

• The Trigonometry Ratios And Angle Tables For The Ratios

A clear explanation for the solutions of the same can be found on the Vedantu website.

7. How many exercises are there in Chapter 9 of Class 10 Maths?

Chapter 9 of Class 10 Maths consists of one exercise that is Exercise 9.1 at the end of this chapter. This section of the chapter has 16 questions that are covered in the first two parts. The questions asked in the exercise are based on the basic concepts of trigonometry and its application, height distance etc. Following this, is a summary of the chapter.

8. How can I score full marks in Class 10 Chapter 9  Maths?