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Class X th Maths NCERT Solutions of  Chapter 9’Some Application of Trigonometry’

Class X th Maths NCERT Solutions of  Chapter 9’Some Application of Trigonometry’created here for helping the students of class 10 in boosting their preparation for term-2 CBSE Board exam 2021-22. Class X th Maths NCERT Solutions of  Chapter 9’Some Application of Trigonometry’is the most important input study material for clearing the basics of Chapter 9’Some Application of Trigonometry’.The questions in chapter 9 ‘Some Application of Trigonometry’are designed by NCERT as per the prescription of CBSE for class 10 students. All the NCERT Solutions of Chapter 9’Some Application of Trigonometry’are completed by a CBSE maths expert teacher in a step by step way therefore we hope every student of class 10 students can understand these solutions.

class 10 maths chapter 9

 

Class X th Maths NCERT Solutions of  Exercise’Some Application of Trigonometry’

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Q.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°.

Solution-

Q1-Chapter 9 class 10

The rope is shown as AC and pole by AB

BC =distance of the rope on the ground from the pole

AC = 20m

 

So, length of pole = 10 m

Class X th Maths NCERT Solutions of  Chapter 9’Some Application of Trigonometry’

Q.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

Solution

 

 

Q2. Chapter 2 class 10

 

Let the tree is parted from point C and its top point D contacts the ground at B.

The broken part DC falls on the ground which is appeared by BC making an angle of 30°with the ground.

The length of the tree,AD =BC+DC= AC +BC ( DC =AC)

So for getting the length of the tree, we have to evaluate AC and BC.

 

………(1)

 

……..(2)

from (1) and (2)  Length of tree

Length of  the tree =8√3 m

Q.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 the elder children she wants to have a steep side 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-

Q3.class 10 chapter 9

The slide for the children below 5 years is shown by the ΔABC and for elder children by the ΔPQR.

So, considering the ΔABC, the length of the slide= AC

sin 30° = AB/AC

1/2 =1.5/AC

AC = 1.5 × 2

AC = 3

Now considering the ΔPQR, the length of the slide= PR

sin 60° = PQ/PR

√3/2 = 3/PR

PR = 6/√3 = 2√3

Therefore the length of the slides for the children below 5 years is 3m and for elder children is 2√3 m.

Q.4: The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower is 30°. Find the height of the tower.

Solution-

Let the height of the tower is h, the tower is supposed to make an angle of 90°,so the Δ formed with the ground will be right Δ.

Q4-chapter 9 class 10

 

The height of the tower will  be = 10√3 m

Q.5: A kite is flying at a height of 60 m 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° . Find the length of the string, assuming that there is no slack in the string.

Solution-

Q5. class 10 chapter 5

 

Length of the string = 40√3 m

Q.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 increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

Solution-

 

Let the boy is standing initially at a distance of y meter from the foot of the building and walked x meter towards building in such a way that the angle of elevation of the top of the tower from his eyes increases from 30°to 60°.

class 10 ,Q6 Chapter 9

 

From the fig.

y = 28.5√3 

The initial distance of the boy from the foot point of the building =28.5√3 m

√3(y–x) = 28.5

√3(28.5√3–x) = 28.5

x√3  = 28.5 ×3–28.5

x√3 = 85.5 –28.5 = 57

x = 19√3

So, the boy walked towards the building 19√3 m

Study 8 th lesson of maths and clear your concepts and doubts on 10 th class cbse trigonometry.

Q.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 20 m high building are 45° and 60°respectively. Find the height of the tower.

Solution-

Q7 class 10 chapter 9 maths

 

Let the height of the transmission tower is h, and the distance of the point from the foot point of the building is x

1 ×x = 20

x =20……..(1)

x√3 = 20 +h……(2)

from (1) and (2)

20√3 –20 = h

h = 20(√3 –1)

So, the height of the transmission tower is 20(√3 –1)

Q.8: A statue, 1.6 m tall, stands on a top of pedestal, from a point on the ground, the angle of elevation of the top of 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-

Let the height of pedestal =h

Q8,Class 10 chapter 9maths

 

Since the pedestal and statue are vertical to the ground making an angle of 90° , so the horizontal distance to the point of observation and the distance to the top of the statue will make a right Δ.

So, from the fig. we have

⇒h = x

x√3 = h +1.6

h√3 –h = 1.6

h(√3–1) = 1.6

h= 0.8(√3 +1)

Therefore, the height of padestal is 0.8(√3 +1) m.

Q.9: The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation 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-

Let the height of the building is = h and the horizontal distance between the building and the tower is = x

Solution-

class 10 maths chapter 10 maths

 

So, from the fig. we have

……….(1)

x = h√3………..(2)

from (1) and (2)

study the simple tricks of solving the roots of the quadratic equation by the complete squire method.

 

Q.10: Two poles of equal heights are standing opposite each other either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30° , respectively. Find the height of poles and the distance of the point from the poles.

Solution-

Q10,class 10 maths chapter 9

 

Let the height of both poles is = h and the distance of one of the pole is x  from the point of observation on the road and of another pole will be = 80–x

So, from the fig. we have

………..(1)

80–x = h√3

………(2)

from (1) and (2) we have

3x = 80–x

x = 20

Therefore the distances of both poles from the point of observation on the road are 20 m and 80–20 =60 

Q11.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 60°. From another point 20 m 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 30° .Find the height of the tower and the width of the canal.

Solution-

Let DC is the  T.V tower,  BC is the width of the canal and B is the point on another bank of the canal and A is another point that is 20 meters away from B.

 Let DC is a tower, B is a point on another bank of canal and A is a point 20 meter away from point B locatd on the same line that is joining the foot of the tower.

Q11 class 10,chapter 9 maths

 

From the fig.

∠A = 30° and ∠DBC = 60°

AC = AB+BC

In ΔBCD

DC = BC√3…. (2)

From equation (i) and (ii)

3BC = 20+BC

BC = 10

Substituting this value of BC in (2) we get

DC = 10√3

Therefore, the hight of tower(DC) is 10√3 m and widh of canal is 10 m.

Q.12: From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

Solution-

Q12,class 10,math chapter 9

 

Let AE is a 7-meter tall building and BD is a cable tower, drawing EC∥AB.

So, AE = BC and CE = AB

In the fig. considering the ΔDCE

DC = CE√3 ………..(1)

CE = BC

BC = 7 (BC = AE)

So, CE = 7, Substituting it in (1)

DC = 7√3 

BD = BC + DC = 7 + 7√3 = 7(√3 +1)

Therefore the height of the tower is 7(√3 +1) m.

 

Q.13: As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45° . If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Solution-

Q13, class 10 chapter 9 math

 

Let A and B are two ships and lighthouse is DC.

AC = 75√3

In the ΔDCB, we have

BC = DC

BC = 75

SO, the distance between two ships(AB) will be= AC –BC

Therefore the distance between ships = 75√3 –75= 75(√3 –1) m

Q.14: A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60° . After some time the angle of elevation reduces to 30°,what will be the distance traveled by the balloon.

 

Solution-

Q12, class 10 chapter 9 maths

 

Let the girl is observing the angle of elevation of 60° when the balloon was on point E and it is reduced to 30°when the balloon reaches D horizontally.

Considering the ΔABE where BE = 88.2 –1.2 = 87 m

…….(1)

From the ΔADC we have

………..(2)

From (1) and (2) we have

BC = AC – AB = 87√3 –87 = 87(√3 –1)

Therefore the distance traveled by the balloon = 87(√3 –1) m

 

Q.15: A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car as an angle of depression of 30°, 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 60° . Find the time taken by the car to reach the foot of the tower from this point.

Solution-

class 10 math chapter 9

 

Let initially the position of the car was A and traveled a distance of x up to B, the height of the tower is h and the distance of the initial position of the car from the point of the tower is y.

In ΔADC, we have

………(1)

In ΔDBC, we have

y√3 –x√3 = h

……(2)

 

3h = h + x√3

2h =x√3

So, the distance traveled by car in 6s is  .

The distance traveled by the car from B to C = y – x = BC is as follows.

BC = y – x

 

Q.16: The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m, from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.

Solution-

 

Let AB is a tower of the height h, C and D are the two points on the ground making the angles of elevation θ and (90 –θ) and located at a distances of 4 m and 9 m respectively from the foot of the tower.

`Q16 class 10 maths chapter 9

 

So, from the fig.

In ΔABC

……….(1)

……….(2)

Multiplying (1) and (2)

h² = 36

h = ±6

Neglecting – sign because height is always positive, so the height of the tower will be 6 m.

Class 10 maths ncert solutions chapter 1: Real numbers

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