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

NCERT solutions for class 10 maths chapter 9 Some Applications of Trigonometry- This is an application chapter of the previous chapter. In this particular chapter, we will study some ways in which trigonometry is used in our daily life. Solutions of NCERT class 10 maths chapter 9 Some Applications of Trigonometry has covered the detailed explanation to each and every question. In the previous class, you have studied trigonometric ratios. Trigonometry is used in navigation, geography, construction of maps, determining the position of an island, etc. In this chapter, we will learn how trigonometry is used for finding the distances and heights and various objects, without actually measuring them. Trigonometry is one of the most ancient topics studied by scholars all over the world. In this chapter, there is only one exercise with 16 questions, based on the practical application of trigonometry. CBSE NCERT solutions for class 10 maths chapter 9 Some Applications of Trigonometry are designed by the subject experts to help students in their preparation of board exams. This chapter introduces new terms like the line of sight, angle of depression and angle of elevation. NCERT solutions for class 10 maths chapter 9 Some Applications of Trigonometry will give you assistance while practicing the questions of the practice exercises including optional exercise. Let's understand a few terms used in this chapter through the diagram.

• The line of sight- It is an imaginary line drawn from the observer's eye to the object viewed by the observer.

• The angle of elevation-  The angle of elevation of an object viewed by an observer, is the angle formed by the line of sight with the horizontal level when the object is above the horizontal level,

• The angle of depression- The angle of depression of an object viewed by an observer, is the angle formed by the line of sight with the horizontal level when the object is below the horizontal level.

Along with this particular chapter, NCERT solutions are available subject wise and chapter wise. If you want to go through them then click on the link given.

## NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Excercise: 9.1

Given that,
The length of the rope (AC) = 20 m. and $\angle ACB$ $=30^o$
Let the height of the pole (AB) be $h$

So, in the right triangle $\Delta ABC$

By using the Sin rule
$\sin \theta = \frac{P}{H} =\frac{AB}{AC}$
$\sin 30^o =\frac{h}{20}$
$h =10$ m.
Hence the height of the pole is 10 m.

Suppose DB is a tree and the AD is the broken height of the tree which touches the ground at C.
GIven that,
$\angle ACB = 30^o$, BC = 8 m
let AB = $x$ m and AD =  $y$
So, AD+AB = DB = $x+y$

In right angle triangle $\Delta ABC$,
$\tan \theta = \frac{P}{B}=\frac{x}{8}$
$\tan 30^o =\frac{x}{8}=\frac{1}{\sqrt{3}}$
So,  the value of $x$ = $8/\sqrt{3}$

Similarily,
$\cos 30^o = \frac{BC}{AC} = \frac{8}{y}$
the value of $y$ is $16/\sqrt{3}$

So, the total height of the tree is-

$x+y=\frac{24}{\sqrt{3}}=8\sqrt{3}$

= 8 (1.732) = 13.856 m (approx)

Suppose $x$ m is the length of slides for children below 5 years and the length of slides for elders children be $y$ m.

Given that,
AF = 1.5 m, BC = 3 m, $\angle AEF = 30^o$ and $\angle BDC = 60^o$

In triangle $\Delta$EAF,
$\sin \theta = \frac{AF}{EF} = \frac{1.5}{x}$
$\sin 30^o = \frac{1.5}{x}$
The value of $x$ is 3 m.

Similarily in $\Delta$CDB,
$\sin 60^o = \frac{3}{y}$
$\frac{\sqrt{3}}{2}= \frac{3}{y}$
the value of $y$ is $2\sqrt{3}$ = 2(1.732) = 3.468

Hence the length of the slide for children below 5 yrs. is 3 m and for the elder children is 3.468 m.

Let the height of the tower AB is $h$ and the angle of elevation from the ground at point C is $\angle ACB = 30^o$
According to question,
In the right triangle $\Delta ABC$,
$\tan \theta = \frac{AB}{BC} = \frac{h}{30}$
$\tan 30^o =\frac{1}{\sqrt{3}}=\frac{h}{30}$
the value of $h$ is $10\sqrt{3}$ = 10(1.732) = 17.32 m
Thus the height of the tower is 17.32 m

A

Given that,
The length of AB = 60 m and the inclination of the string with the ground at point C is $\angle ACB = 60^o$.
Let the length of the string AC be $l$.
According to question,
In right triangle $\Delta$CBA,
$\sin 60^{o} = \frac{AB}{AC} = \frac{60}{l}$
$\frac{\sqrt{3}}{2} = \frac{60}{l}$
The value of length of the string ($l$)  is $40\sqrt{3}$ = 40(1.732) = 69.28 m
Hence the length of the string is 69.28 m.

Given that,
The height of the tallboy (DC) is 1.5 m and the height of the building (AB) is 30 m.
$\angle ADF = 30^o$ and $\angle AEF = 60^o$

According to question,
In right triangle AFD,
$\\\Rightarrow \tan 30^o=\frac{AF}{DF} = \frac{28.5}{DF}\\\\\Rightarrow \frac{1}{\sqrt{3}} = \frac{28.5}{DF}$
So, DF = $(28.5)\sqrt{3}$

In right angle triangle $\Delta AFE$,
$\tan 60^o =\frac{AF}{FE}=\frac{28.5}{EF}$
$\sqrt{3}=\frac{28.5}{EF}$
EF = $9.5\sqrt{3}$

So, distance walked by the boy towards the building = DF - EF = $19\sqrt{3}$

Suppose BC = $h$ is the height of transmission tower and the AB be the height of the building and  AD is the distance between the building and the observer point (D).
We have,
AB = 20 m, BC = $h$ m and AD = $x$ m
$\angle CDA = 60^o$ and $\angle BDA = 45^o$

According to question,
In triangle $\Delta$ BDA,
$\tan 45^o = \frac{AB}{AD}=\frac{20}{x}$
So, $x$ = 20 m

Again,
In triangle $\Delta$CAD,

$\\\Rightarrow \tan 60^o = \frac{AB+BC}{AD}=\frac{20+h}{20}\\\\\Rightarrow \sqrt{3}= 1+\frac{h}{20}\\\\\Rightarrow h=20(\sqrt{3}-1)\\\\\Rightarrow 20(0.732) = 14.64 m$

Answer- the height of the tower is 14.64 m

Let the height of the pedestal be $h$ m. and the height of the statue is 1.6 m.
the angle of elevation of the top of the statue and top of the pedestal is($\angle DCB=$ $60^o$ )and($\angle ACB=$ $45^o$) respectively.

Now,
In triangle $\Delta ABC$,
$\tan 45^o =1 =\frac{AB}{BC}=\frac{h}{BC}$
therefore,  BC = $h$ m

In triangle $\Delta CBD$,
$\\\Rightarrow \tan 60^o = \frac{BD}{BC}=\frac{h+1.6}{h}\\\\\Rightarrow \sqrt{3}= 1+\frac{1.6}{h}$
the value of  $h$ is $0.8(\sqrt{3}+1)$ m
Hence the height of the pedestal is  $0.8(\sqrt{3}+1)$m

It is given that, the height of the tower (AB) is 50 m. $\angle AQB = 30^o$ and $\angle PBQ = 60^o$
Let the height of the building be $h$ m

According to question,
In triangle PBQ,
$\tan 60^o = \frac{PQ}{BQ} = \frac{50}{BQ}$
$\\\sqrt{3} = \frac{50}{BQ}\\ BQ = \frac{50}{\sqrt{3}}$.......................(i)

In triangle ABQ,

$\tan 30^o = \frac{h}{BQ}$
${BQ}=h\sqrt{3}$.........................(ii)
On equating the eq(i) and (ii) we get,

$\frac{50}{\sqrt{3}}=h\sqrt{3}$
therefore, $h$ = 50/3 = 16.66 m = height of the building.

Given that,
The height of both poles are equal  DC = AB. The angle of elevation of of the top of the poles are $\angle DEC=30^o$ and $\angle AEB=60^o$ resp.
Let the height of the poles be  $h$ m and CE = $x$ and BE = 80 - $x$

According to question,
In triangle DEC,

$\\\Rightarrow \tan 30^o = \frac{DC}{CE} = \frac{h}{x}\\\\\Rightarrow \frac{1}{\sqrt{3}}= \frac{h}{x}\\\\\Rightarrow x=h\sqrt{3}$..............(i)

In triangle AEB,
$\\\Rightarrow \tan 60^o = \frac{AB}{BE}=\frac{h}{80-x}\\\\\Rightarrow \sqrt{3}=\frac{h}{80-x}\\\\\Rightarrow x=80 - \frac{h}{\sqrt{3}}$..................(ii)
On equating eq (i) and eq (ii), we get

$\sqrt{3}h=80 - \frac{h}{\sqrt{3}}$
$\frac{h}{\sqrt{3}}=20$
$h=20\sqrt{3}$ m
So, $x$ = 60 m

Hence the height of both poles is ($h=20\sqrt{3}$)m and the position of the point is at 60 m from the pole CD and 20 m from the pole AB.

Suppose the $h$ is the height of the tower AB and BC = $x$ m
It is given that, the width of CD is 20 m,
According to question,

In triangle $\Delta ADB$,
$\\\Rightarrow \tan 30^o = \frac{AB}{20+x}=\frac{h}{20+x}\\\\\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{20+x}\\\\\Rightarrow 20+x = h\sqrt{3}\\\\\Rightarrow x = h\sqrt{3}-20$............(i)

In triangle ACB,
$\\\Rightarrow \tan 60^o = \frac{h}{x}=\sqrt{3}\\\\\Rightarrow x= \frac{h}{\sqrt{3}}$.............(ii)

On equating eq (i) and (ii) we get:

$h\sqrt{3}-20= \frac{h}{\sqrt{3}}$
from here we can calculate the value of  $h=10\sqrt{3}= 10 (1.732) = 17.32\: m$  and the width of the canal is 10 m.

Let the height of the cable tower be (AB = $h+7$)m
Given,
The height of the building is 7 m and angle of elevation of the top of the tower $\angle ACE = 60^o$ , angle of depression of its foot $\angle BCE = 45^o$.

According to question,

In triangle $\Delta DBC$,
$\\\tan 45^o = \frac{CD}{BD} = \frac{7}{BD} = 1\\ BD =7\ m$
since DB = CE = 7 m

In triangle $\Delta ACE$,

$\\\tan 60^o = \frac{h}{CE}=\frac{h}{7}=\sqrt{3}\\ \therefore h = 7\sqrt{3}\ m$

Thus, the total height of the tower equal to $h+7$ $=7(1+\sqrt{3}\) m$

Given that,
The height of the lighthouse (AB) is 75 m from the sea level. And the angle of depression of two different ships are $\angle ADB = 30^0$ and $\angle ACB = 45^0$ respectively

Let the distance between both the ships be $x$ m.
According to question,

In triangle $\Delta ADB$,

$\tan 30^0 = \frac{AB}{BD}=\frac{75}{x+y} = \frac{1}{\sqrt{3}}$
$\therefore x+y = 75 \sqrt{3}$.............(i)

In triangle $\Delta ACB$,

$\tan 45^0 = 1 =\frac{75}{BC}=\frac{75}{y}$
$\therefore y =75\ m$.............(ii)

From equation (i) and (ii) we get;
$x = 75(\sqrt{3}-1)=75(0.732)$
$x = 54.9\simeq 55\ m$

Hence, the distance between the two ships is approx 55 m.

Given that,
The height of the girl is 1.2 m. The height of the balloon from the ground is 88.2 m and the angle of elevation of the balloon from the eye of the girl at any instant is ($\angle ACB =60^0$) and after some time $\angle DCE =30^0$.

Let the $x$ distance travelled by the balloon from position A to position D during the interval.
AB = ED = 88.2 - 1.2 =87 m

Now, In triangle $\Delta BCA$,

$\\\tan 60^0 = \sqrt{3}=\frac{AB}{BC}=\frac{87}{BC}\\ \therefore BC = 29\sqrt{3}$

In triangle $\Delta DCE$,

$\\\tan 30^0 = \frac{1}{\sqrt{3}}=\frac{DE}{CE}=\frac{87}{CE}\\ \therefore CE = 87\sqrt{3}$

Thus, distance traveled by the balloon from position A to D

$= CE - BC =87\sqrt{3}-29\sqrt{3} = 58\sqrt{3}$ m