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

May 22, 2023

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NCERT Solutions Class 10 Maths Chapter 9– Some Applications of Trigonometry have actually been established by the group at GFG as a tool to help trainees in solving concerns associated with applications of Trigonometry. All the issues in this chapter’s workout from the NCERT book are covered in the NCERT Solutions for Class 10 Mathematics.

The research study of how trigonometry is utilized almost in life is the topic of this chapter. The line of sight, angle of variance, angle of elevation, and angle of anxiety are all covered in this subject. The chapter’s workout is based upon the conversation of how trigonometry might be utilized to figure out the heights and ranges of numerous things.

These Solutions cover all 4 workouts of the NCERT Class 10 Maths Chapter 9, which are as follows:

Class 10 Maths NCERT Solutions Chapter 9 Workouts
NCERT Maths Solutions Class 10 Workout 9.1 Set 1, Set 2 — 16 Concerns (16 Long Responses)

Some Application of Trigonometry– Workout 9.1

Concern 1. A circus artist is climbing up a 20 m long rope, which is securely extended and connected from the top of a vertical pole to the ground. Discover the height of the pole, if the angle made by the rope with the ground level is 30 ° (see Fig.).

Service:

In rt ∆ ABC,

AB = pole =?

AIR CONDITIONER = rope = 20m

sinθ = $frac{P}{H}$

sin30 ° = $frac{AB}{AC}$

AB = 1/2 * 20

AB = 10m

Height of pole = 10m

Concern 2. A tree breaks due to storm and the damaged part flexes so that the top of the tree touches the ground making an angle 30 ° with it. The range in between the foot of the tree to the point where the leading touches the ground is 8 m. Discover the height of the tree.

Service:

In rt ∆ ABC,

BC = 8m

$frac{P}{B} = frac{AB}{BC}$ = tan30 °

$frac{AB}{BC}$ = 1/ √ 3

AB= 8/ √ 3-( 1)

Now,

$frac{BC}{ AC}$ = cos30 °

8/AC= √ 3/2

√ 3AC = 16

AIR CONDITIONER =16/ √ 3 -( 2)

From (1) and (2 )(* )Height of tree= AB + AIR CONDITIONER

= 8/ √ 3 * 16 √ 3

= 8 √ 3 m(* )8 * 1.73= 13.84 m

The height of the tree is 13.84

Concern 3. A professional prepares to set up 2 slides for the kids to play in a park. For the kids listed below the age of 5 years, she chooses 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 senior kids, she wishes to have a high 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?

Service:

In rt ∆ ABC,

AB = 1.5 m

AIR CONDITIONER = side =?

= sin30 °

1.5/ AIR CONDITIONER = 1/2

$frac{P}{H} = frac{AB}{AC}$ AIR CONDITIONER =1/5 * 2

AIR CONDITIONER = 3m

In rt ∆ PQR,

PQ = 3m

PR = side =?

= sin60 °

3/PR= √ 3/2

$frac{P}{H} = frac{PQ}{PR}$ √ 3 PR= 6

PR= 6/ √ 3

6/ √ 3 * √ 3/ √ 3

= 2 √ 3(* )= 2 * 1.73

= 3.46 m

Concern 4. The angle of elevation of the top of a tower from a point on the ground, which is 30 m far from the foot of the tower, is 30 °. Discover the height of the tower.

Service:

In rt ∆ ABC,

AB = tower=?

BC = 30m

= tan30 °

AB/30= 1/ √ 3

AB= 30/ √ 3

$frac{P}{B} = frac{AB}{BC}$ AB= 30/ √ 3 * √ 3/ √ 3

=( 30 √ 3 )/ 3= 10 √ 3(* )= 10 * 1.73

= 17.3 m(* )The height of tower 17.3 m

Concern 5. A kite is flying at a height of 60 m in the air. The string connected to the kite is briefly connected to a point on the ground. The disposition of the string with the ground is 60 °. Discover the length of the string, presuming that there is no slack in the string.

Service:

In rt ∆ ABC,

AB = 6Om

AIR CONDITIONER= string =?

= sin60 °

60/AC = √ 3/3

√ 3 AIR CONDITIONER =60 * 2

AIR CONDITIONER= 120/120/( √ 3) * √ 3/ √ 3

120/ √ 3 * √ 3/ √ 3

$frac{P}{H} = frac{AB}{AC}$ 40 =√ 3

40 * 1.73= 69.20 m

Length of the string is 69.20 m

Concern 6. A 1.5 m high young boy is standing at some range from a 30 m high structure. The angle of elevation from his eyes to the top of the structure increases from 30 ° to 60 ° as he strolls towards the structure. Discover the range he strolled towards the structure.

Service:

In fig AB= AE– 1.5

= 30– 1.5

= 28.5

In rt ∆ ABD,

= tan30 °

= 28.5/ BD= 1/ √ 3

BD = 28.5 √ 3-( 1)

In rt ∆ ABC,

= tan60 °

$frac{P}{B} = frac{AB}{BD}$ 28.5/ BC * √ 3

√ 3 BC = 28.5

BC= 28.5/ √ 3 – (2)

CD = BD − BC

$frac{P}{B} = frac{AB}{BC}$ = 28.5 √ 3– 28.5/ √ 3

= 28.5( 2/ √ 3)

57/ √ 3 * √ 3/ √ 3=( 57 √ 3)/ 3= 19 √ 3

19 * 1.73= 32.87 m

The young boy strolled 32.87 m towards the structure.

Concern 7. From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower repaired at the top of a 20 m high structure are 45 ° and 60 ° respectively. Discover the height of the tower.

Service:

In fig:

AB = tower =?

BC= structure = 20m

In rt ∆ BCD (* ) = tan45 °

20/CD= 1/1

CD= 20

In rt. ∆ ACD,

= tan60 °

AC/20= √ 3/1

$frac{P}{B} = frac{BC}{CD}$ AIR CONDITIONER= 20 √ 3- (1)(* )AB= AC-BC

20 √ 3– 20

20( √ 3– 1)

20( 1.732– 1)

$frac{P}{B} = frac{AC}{CD}$ 20 (0.732 )

14.64 m(* )The height of the tower is 14.6 m

Concern 8. A statue, 1.6 m high, bases on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60 ° and from the very same point, the angle of elevation of the top of the pedestal is 45 °. Discover the height of the pedestal.

Service:

In fig: AB= statue= 1.6 m

BC= pedestal =?

In rt ∆ ACD

= tan60 °

1.6+ BC/CD = √ 3

√ 3 CD = 1.6 +BC(* )CD = 1.6 +BC/ √ 3-( 1)

In rt ∆ BCD,

= tan45 °

= 1/1

CD =BC

From( 1)

$frac{P}{B} = frac{AC}{CD}$ 1.6+ BC/ √ 3= BC/1

√ 3 BC = 1.6 +BC

1.732 BC– 1 BC =1.6

0.732 * BC =1.6(* )BC= 1.6/ 0.732 (* )BC= 16/10 * 100/732= 1600/732

BC =2.18 m

$frac{BC}{CD}$ Height of pedestal is 2.18 m

$frac{BC}{CD}$ Concern 9. The angle of elevation of the top of a structure from the foot of the tower is 30 ° and the angle of elevation of the top of the tower from the foot of the structure is 60 °. If the tower is 50 m high, discover the height of the structure.

Service:

In fig:

AB = tower = 50m

DC = structure = ?

In rt. ∆ ABC,

= tan60 °

√ 3 BC =50(* )BC= 50/ √ 3

In rt. ∆ DCB

= tan30 °

= 1/ √ 3

DC= 50/ √ 3

DC= 50/ √ 3 * 1/ √ 3

DC= 50/3(* )DC =(* )The height of the structure is

m(* )Concern 10. 2 poles of equivalent heights are standing opposite each other on either side of the roadway, which is 80 m large. From a point in between them on the roadway, the angles of elevation of the top of the poles are 60 ° and 30 °, respectively. Discover the height of the poles and the ranges of the point from the poles.

Service:

AB and CD on equivalent poles.

$frac{P}{B} = frac{AB}{BC}$ Let their height = h

Let DP= x

Then PB= BD– x

In rt. ∆ CDP,

$frac{P}{B} = frac{DC}{BC}$ = tan60 °

$frac{P}{B} = frac{DC}{frac{50}{√3}}$ h/x= √ 3/1

h= √ 3 x – (1)

In rt. ∆ ABP

= tan30 °

h/( 80– x) = 1/ √ 3 $16tfrac{2}{3}$

h=( 80– x)/ √ 3-( 2) $16tfrac{2}{3}$ From( 1 )and( 2)

( √ 3 x)/ 1= 80– x/ √ 3

3x =80– x(* )3x+ x= 80

4x =80

X= 80/4

X =20

Putting worths of X in formula 1

h= √ 3 x

$frac{P}{B} = frac{CD}{DP}$ h= √ 3( 20)

h =1.732( 20 )(* )h= 34.640

Height of each pole =34.64 m

The point is 20m far from very first pole and 60m far from 2nd pole.

$frac{P}{B} = frac{AB}{BP}$ Concern 11. A television tower stands vertically on a bank of a canal. From a point on the other bank straight opposite the tower, the angle of elevation of the top of the tower is 60 °. From another point 20 m far from this point on the line joining this indicate the foot of the tower, the angle of elevation of the top of the tower is 30 °( see Fig.). Discover the height of the tower and the width of the canal. (* ) Service:

In fig: AB= tower=?

CB = canal =?

In rt. ∆ ABC,

tan60 ° =

h/x = √ 3

h = √ 3 x -( 1)

In rt. ∆ ABD

= tan 30 °

=1/ √ 3

h=( x +20)/ √ 3 -( 2 )

From( 1) and( 2) (* )√ 3/1=( x+ 20)/ √ 3

3x =x+ 20

3x– x= 20

2x = 20

X= 20/2

X= 10

Width of the canal is 10m

Putting worth of x in formula 1

h= √ 3 x(* )= 1.732 (10)

=17.32

Height of the tower 17.32 m. (* ) Concern 12. From the top of a 7 m high structure, the angle of elevation of the top of a cable television tower is 60 ° and the angle of anxiety of its foot is 45 °. Figure out the height of the tower.

Service: $frac{AB}{BC}$

In fig: ED= structure= 7m

AIR CONDITIONER = cable television tower =?

In rt ∆ EDC,

$frac{AB}{BD}$ = tan45 °

$frac{h}{x + 20}$ 7/x= 1/1

DC =7

Now, EB= DC = 7m

In rt. ∆ ABE,

= tan60 °(* )AB/7= √ 3/1(* )Height of tower =AIR CONDITIONER= AB + BC(* )7 √ 3 + 7

= 7( √ 3 + 1 )(* )= 7( 1.732 + 1)

= 7( 2.732)

Height of cable television tower =19.125 m

Concern 13. As observed from the top of a 75 m high lighthouse from the sea-level, the angles of anxiety of 2 ships are 30 ° and 45 °. If one ship is precisely behind the other on the very same side of the lighthouse, discover the range in between the 2 ships.

Service:

In fig:

AB = lighthouse =75m

D and C are 2 ships

DC =?

In rt. ∆ ABD,

= tan30 °

75/BD= 1/ √ 3

BD= 75 √ 3

In rt. ∆ ABC(* )= tan45 °

75/BC =1/1(* )BC= 75

$frac{ED}{DC}$ DC= BD– BC

= 75 √ 3– 75(* )75( √ 3– 1)

75( 1.372– 1)

34.900

For this reason, range in between 2 sheep is 34.900

$frac{AB}{BC}$ Concern 14. A 1.2 m high woman finds a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the woman at any immediate is 60 °. After a long time, the angle of elevation lowers to 30 ° (see Fig.). Discover the range taken a trip by the balloon throughout the period.

Service:

In fig: AB = AIR CONDITIONER– BC

= 88.2– 1.2

= 81m

In rt. ∆ ABE

= 87/EB = tan30 °

87/EB = 1/ √ 3

EB = 87 √ 3

In rt. ∆ FDE

= tan60 °

√ 3 ED = 87

ED= 87/ √ 3

DB= DB– ED

87 √ 3– 87/ √ 3

$frac{AB}{BD}$ 87( √ 3– 1/ √ 3)(* )= 87( 3– 1/ √ 3)(* )= 87( 2/ √ 3) =174/ √ 3 * √ 3/ √ 3

= 174 * √ 3/3= 58 √ 3

58 * 1.732= 100.456 m

Range taken a trip by balloon is 100.456 m(* )Concern 15. A straight highway results in the foot of a tower. A male standing at the top of the tower observes a vehicle at an angle of anxiety of 30 °, which is approaching the foot of the tower with a consistent speed. 6 seconds later on, the angle of anxiety of the cars and truck is discovered to be 60 °. Discover the time taken by the cars and truck to reach the foot of the tower from this point.

$frac{AB}{BC}$ Service:

In fig: AB is tower

In rt. ∆ ABD

= tan30 °

= 1/ √ 3(* ) DB = √ 3 AB -( 1)

In rt. ∆ ABC

= tan60 °

BC = AB/ √ 3-( 2)

DC= DB– BC (* )= √ 3 AB– AB/ √ 3

AB( 3– 1/ √ 3)

CD= 2AB/ √ 3

S1= S2

frac {D1} {T1} =frac {D2} {T2}

2/ √ 3AB/6 = AB/ √ 3/t

2t= 6

$frac{AB}{EB}$ t =6/2

t= 3sec(* )Concern 16. The angles of elevation of the top of a tower from 2 points at a range of 4 m and 9 m from the base of the tower and in the very same straight line with it are complementary. Show that the height of the tower is 6 m.

Service:

In fig: AB is tower

$frac{FD}{ED}$ To show:

AB = 6m

Offered:

BC = 4m DB = 9m

In ∆ ABC

= tanθ

AB/4 = tanθ -( 1 )

In ∆ ABD

= tan (90 °- θ)

AB/9 = 1/ tanθ

9/AB = tanθ -( 2 )

From (1) and (2 )

AB/4 = 9/AB

AB

2

$frac{AB}{DB}$ = 36

$frac{AB}{DB}$ AB = √ 36

AB = √( 6 * 6)

AB = 6m

$frac{AB}{BC}$ Height of the tower is 6m.

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Frequently Asked Questions on NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry

$Speed = frac{Distance}{Time}$

Q1: Why is it essential to discover the applications of trigonometry?

Response:

$frac{DC}{6} = frac{CB}{t}$

It is essential since topics such as astronomy and physics utilize the applications of trigonometry, to determine the separation in between worlds and stars, identify their speed, and evaluate waves.

Q2: What subjects are covered in NCERT Solutions for Class 10 Maths Chapter 9– Some Applications of Trigonometry?

Response: