Some Applications of Trigonometry

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Applications of Trigonometry Class 10 Maths NCERT Solutions Chapter 9 is concerned with the understanding of various aspects of why, where and how trigonometry is used. Students preparing for some applications of trigonometry Class 10 will be able to clear all their concepts on applications of trigonometry at the root level. Trigonometry is used to refine complicated problems in many physical sciences and engineering topics today. Expert faculty of Toppr produced these solutions to assist students with their first term exam preparations. It covers all major concepts in detail, allowing students to understand the ideas better.

NCERT Solutions Applications of Trigonometry class 10 Maths chapter 9 concentrate on the essential concepts such as the angle of elevation, angle of depression, and angle of sight. All of these solutions are designed with the new CBSE pattern in mind so that students have a complete understanding of their tests.

Some Applications of Trigonometry chapter 9 maths class 10 Questions and Answers are very useful for getting good grades in tests and properly preparing you with all of the important concepts. These NCERT Solutions are valuable tools that can assist you not only in covering the full syllabus but also in providing an in-depth analysis of the subjects. The Class 10 Maths NCERT Solutions Chapter 9 are available in pdf format below, and some of them are also included in the exercises.

Table of Content

Exercise 9.1

Question 1

A circus artist is climbing a 20 m 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 30$_{o}$.

Solution

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Let AB be the vertical pole and CA be the rope. Then,

$∠ACB=30_{o}$ and $AC=20m$

In right $△$ ABC,

$sin30_{o}=ACAB $

$21 =20AB $

$AB=10m$

Therefore, the height of the pole is 10 m.

Question 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 of $30_{o}$ with the ground. The distance between the foot of the tree to the point where the top touches the ground is $8$ m. Find the height of the tree.

Solution

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Let the Height of the Tree $=AB+AD$

Now, when it breaks a part of it will remain perpendicular to the ground $(AB)$ and remaining part $(AD)$ will make an angle of $30_{o}$

Now, in $△ABD$

$cos30_{o}=ADBD ⇒BD=23 AD ⇒AD=3 2×8 $

and given that $BD=8m$

Now, in $△ABD$

$cos30_{o}=ADBD ⇒BD=23 AD ⇒AD=3 2×8 $

also, in the same Triangle

$tan30_{o}=BDAB ⇒AB=3 8 $

$∴$ Height of tree $=AB+AD=(3 16 +3 8 )m=3 24 m=83 m$

Question 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 $1.5$ m, and is inclined at an angle of $30_{∘}$ to the ground, whereas for elder children, she wants to have a steep slide at a height of $3$m, and inclined at an angle of $60_{∘}$ to the ground. What should be the length of the slide in each case?

Solution

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In $△ABC$

$sin30_{o}=ACBC $

$21 =AC1.5 $

$∴AC=3m$

Elder children

$XZYZ =23 =XZ3 $

$XZ=3 6 =23 m$

Question 4

The angle of elevation of the top of a tower from a point on the ground, which is $30m$ away from the foot of the tower, is $30_{∘}$. Find the height of the tower.

Solution

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In $△PQR,$

$tan30_{∘}=PQPR $

$⇒PR=PQtan30_{∘}$

$⇒PR=3 30 $

$⇒PR=103 m$

$∴$ Height of the tower is $103 m$.

Question 5

A kite is flying at a height of $60m$ 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 $60_{o}$. Find the length of the string assuming that there is no slack in the spring.

Solution

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$P$ is the position of the kite. Its height from the point $Q$ (on the ground)$=PQ=60m$

Let $OP=l$ be the length of the string

Let $OP=l$ be the length of the string

$∠POQ=60_{o}$ (given)

$sin(Θ)=HypotenuseOpposite $

Now, $OPPQ =sin60_{o}$

$⇒$ $l60 =sin60_{o}=23 $

$⇒$ $l60 =23 $

$⇒$ $l=3 3×2×20 m$

$⇒$ $l=403 m$

The length of the string is $403 m$

Question 6

A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increase from $30_{∘}$ to $60_{∘}$ as he walks towards the building. Find the distance he walked towards the building.

Solution

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REF.Image

AB= 30-1.5=28.5 cm

In $ΔABD:−tan30_{∘}=BDAB $

$⇒BD=(28.5)3 $

In $ΔABC:−tan60_{∘}=BCAB $

$⇒BC=3 (28.5) $

CD= BD-BC

$=28.5(3 −3 1 )$

$=28.5(3 2 )=32.91m$

Question 7

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

Solution

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Let DC be the tower and BC be the building. Then,

$∠CAB=45_{o},∠DAB=60_{o},BC=20m$

Let height of the tower, $DC=hm$.

In right $△ABC$,

$tan45_{o}=ABBC $

$1=AB20 $

$AB=20m$

In right $△ABD$,

$tan60_{o}=ABBD $

$3 =20h+20 $

$h=20(3 −1)m$

Question 8

A statue 1.6 m tall stands on the top of pedestal. From a point on the elevation of the top of the statue is 60 and from the same point the angle of elevation of the top of the pedestal is 45. Find the height of the pedestal.

Solution

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In $ΔBCD$

$tan30_{o}=CDBC $

$3 1 =CDh $

$CD=3 h$m ........$(1)$

Now,

In $ΔACD$

$tan60_{o}=CDAC $

$=CDAB+BC $

From equation $(1)$ substitution value of CD

$tan60_{o}=3 h1.6+h $

$⇒h3 ×3 =1.6+h$

$⇒3h−h=1.6$

$2h=1.6$

$h=0.8$m

Height of Pedestal.

Question 9

The angle of elevation of the top of the building from the foot of the tower is $30$ and the angle of the top of the tower from the foot of the building is $60$. If the tower is $50$ m high, find the height of the building.

Solution

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Given height of tower $CD=50m$

Let the height of the building, $AB=h$

In right angled $△BDC$,

$⇒$ $tan60_{o}=BDCD $

$⇒$ $3 =BD50 $

$⇒$ $BD=3 50 m$

In right angled $△ABD$,

$⇒$ $tan30_{o}=BDAB $

$⇒$ $3 1 =3 50 h $

$∴$ $3 1 =503 h $

$∴$ $h=350 =16.66m$

$∴$ The height of the building is $16.66m$.

Question 10

Two poles of equal heights are standing opposite to each other, on either side of the road, which is $80m$ wide. From a point between them on the road, the angles of elevation of top of the poles are $60_{o}$ and $30_{o}$ respectively. Find the height of the poles.

Solution

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Given that:

$∠APB=60_{∘},∠CPD=30_{∘},AC=80m$

To find:

The height of the pole$=AB=CD=?$

Solution:

Let $AB$ and $CD$ be the two poles of equal height and $P$ be the point on the road between the poles.

In $△APB,$

$tan60_{∘}=APAB $

or, $AP=AB×tan60_{∘}1 $

or, $AP=3 AB $ $−−−−−−−(i)$

In $△PCD,$

$tan30_{∘}=CPCD $

or, $CP=CD×tan30_{∘}1 $

or, $CP=3 CD=3 AB$ $∵AB=CD$ $−−−−−−−(ii)$

Adding eqn. (i) and eqn. (ii) we get,

$AP+CP=3 AB +AB3 $

or, $AC=AB(3 +3 1 )$

or, $80m=43 AB $

or, $AB=203 m$

Therefore, height of the pole$=203 m=34.64m$

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Trigonometry is the branch of mathematics that deals with the relationships between the sides and angles of triangles, as well as the related functions of triangle angles. Trigonometry was developed in response to a requirement in astronomy. Since then, astronomers have used it to measure distances from Earth to planets and stars, among other things. In geography and navigation, trigonometry is also used. Trigonometry is used to create maps and determine the position of an island in relation to longitudes and latitudes.

The line of sight in the illustration above refers to the line drawn from an observer's eye to the top of the object. The observer is looking up at the object's top. The angle θ, created by the line of sight with the horizontal line is known as the angle of elevation of the top of an object from the observer's eye.

The line AC in the above picture is the line of sight when the observer looks downwards from the top of the building at A to the object at C. Angle of depression is defined as the angle produced by the line of sight with the horizontal when the observer lowers his/her head.

Trigonometric ratios can be used to calculate the height or length of an object, as well as the distance between two distant objects.

Related Chapters

- Chapter 1 : Real NumbersChapter 2 : PolynomialsChapter 3 : Pair of Linear Equations in Two VariablesChapter 4 : Quadratic EquationsChapter 5 : Arithmetic ProgressionChapter 6 : TrianglesChapter 7 : Coordinate GeometryChapter 8 : Introduction to TrigonometryChapter 10 : CirclesChapter 11 : ConstructionsChapter 12 : Area Related to CirclesChapter 13 : Surface Areas and VolumesChapter 14 : StatisticsChapter 15 : Probability

Q1. What are the most important topics in NCERT Class 10 Chapter 9 Some Applications of Trigonometry?

Answer: NCERT Solutions Class 10 Maths Chapter 9's main concepts Trigonometry applications include the angle of depression, angle of elevation, and line of sight. All questions based on these themes have detailed step-by-step responses. Students should make a point of drawing the corresponding figures after reading the question to better comprehend these topics. On the Toppr website, there is a comprehensive explanation for the answers.

Q2. How many exercises are there in Chapter 9 Some Applications of Trigonometry of Class 10 Maths?

Answer: Exercise 9.1 at the end of Chapter 9 Some Applications of Trigonometry of Class 10 Maths is the only exercise in this chapter. This chapter's portion has 16 questions that are covered in the first two parts. There are a total of 16 questions in Class 10 Maths Chapter 9 Some Applications of Trigonometry, with 6 being quite easy, 5 being moderate, and 5 being long answer type problems.

Q3. Explain a few real-life applications of trigonometry?

Answer: Today, trigonometry is utilized to refine difficult problems in many physical sciences and engineering topics. Trigonometry is employed in the development of computer music because sound travels in the form of waves, and this wave pattern employing sine and cosine functions aids in the development of computer music. Analytical trigonometry is essential in engineering fields such as mechanical engineering, electronics, and mechatronics. It has been used by astronomers to calculate distances from Earth to planets and stars, among other things. Trigonometry is also employed in geography and navigation. Trigonometry is used to make maps and calculate an island's position in reference to longitudes and latitudes. It is used in the marine and aviation industries. It is used in cartography (creation of maps). Trigonometry is also employed in satellite systems.

Related Chapters

- Chapter 1 : Real NumbersChapter 2 : PolynomialsChapter 3 : Pair of Linear Equations in Two VariablesChapter 4 : Quadratic EquationsChapter 5 : Arithmetic ProgressionChapter 6 : TrianglesChapter 7 : Coordinate GeometryChapter 8 : Introduction to TrigonometryChapter 10 : CirclesChapter 11 : ConstructionsChapter 12 : Area Related to CirclesChapter 13 : Surface Areas and VolumesChapter 14 : StatisticsChapter 15 : Probability