NCERT Solutions Class 10 Maths Chapter 9 Some Applications of Trigonometry Exercise 9.1

Class 10 NCERT Solutions Maths Chapter 9 Exercise 9.1 will take you through the most valuable applications of trigonometry. Trigonometry is one of the most crucial topics in mathematics, which helps solve problems related to angles and sides of right-angled triangles and calculates distances in real-life situations.

The solutions are designed according to the latest CBSE syllabus and exam pattern to help you succeed in your final exams. So, let’s dive deeper into this fascinating chapter of mathematics. 

1.0NCERT Solutions Class 10 Maths Chapter 9 Some Applications of Trigonometry Exercise 9.1 : Free PDF

2.0Introduction to Applications of Trigonometry

The use of trigonometry dates back to ancient times, especially in astronomy and navigation. Trigonometry was utilized by astronomers to compute distances between stars and planets. As time went on, trigonometric concepts made their way into other disciplines, like geography and cartography, and it aided in determining the relative positions of objects, including islands, cities, and landmarks, according to latitude and longitude. In this chapter, we discuss its use in determining the heights and distances of objects that cannot be directly measured. In this exercise, we will use the trigonometric ratios to calculate these distances.     

3.0Class 10 Maths Chapter 9 Exercise Overview: Key Concepts

The Angle of Elevation

The exercise predominantly uses the concept of angle of elevation to solve its many questions. The angle of elevation is the angle created by the line of sight when an observer views an object above the horizontal level looking upwards. In other words, the angle of elevation is formed when a person looks upward at an object (mostly on top of the building). For instance, take a student viewing the top of a minar (tower). The angle between the line of sight from the student's eyes to the top of the minar and the ground is the angle of elevation.

The Angle of Depression

Apart from the angle of elevation, Exercise 9.1 also includes another important concept, the angle of depression. The angle of depression refers to the angle formed when an observer looks at an object that is below the level of their horizontal eye line. In simple words, an angle of depression forms when an observer looks downward at an object. For instance, take a girl sitting on a balcony who is viewing a flower pot resting on a step below. The angle between the line of sight from her eyes to the pot and the horizontal line is the angle of depression.

Using Trigonometric Ratios to Calculate Height and Distances

The main objective of this exercise is to calculate the height and distance of an object, like a tall building, using trigonometric ratios. The trigonometric ratios include sine, cosine, tangent, secant, cosecant, and cotangent. However, in this exercise, the most used trigonometric ratios are sine, cosine, and tangent. These ratios are used to either find the height and distance of an object or the angle of elevation or depression formed by the observer. The trigonometric ratios can also be written as: 

$\sin \theta =\frac{\text{Perpendicular}}{\text{Hypotenuse}}$

$\cos \theta =\frac{\text{Base}}{\text{Hypotenuse}}$

$\tan \theta =\frac{\text{Perpendicular}}{\text{Base}}$

According to the exercise: 

• $\theta$ = The angle of elevation or depression.
• Perpendicular = Height of an object to be observed. 
• Hypotenuse = Shortest distance from the observer to the top or bottom of the object. 
• Base = Horizontal distance between the object and the observer. 

Unlock the potential of trigonometry and elevate your problem-solving ability to new dimensions with NCERT Solutions Class 10 Maths : Chapter 9 Exercise 9.1 – where each angle is a step towards mastering the application of trigonometry. 

4.0NCERT Solutions Class 10 Maths Chapter 9 Some Applications of Trigonometry Exercise 9.1 : Detailed Solutions

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°. (see fig.).

Solution:

AC = 20 m is the length of the rope.

Let AB = h meters be the height of the pole.

∠ACB = 30° (Given)

Now,

AB/AC = sin 30° = 1/2

=> h/20 = 1/2

=> h = 10 m

∴ the height of the pole is 10 m.

2. A tree breaks due to a storm and the broken part bends so that the top of the tree touches the ground making an angle of 30° 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:

Let a tree be broken at point A, and its top be touching the ground at point B.

Now, in right triangle AOB, we have:

AO/OB = tan 30°

=> AO/8 = 1/√3

=> AO = 8/√3 meters

Also, AB/OB = sec 30°

=> AB/8 = 2/√3

=> AB = (2 * 8)/√3 = 16/√3 meters

Now, the height of the tree OP = OA + AP

= OA + AB

= 8/√3 + 16/√3 = 24/√3 * (√3/√3) meters = 8√3 meters

Therefore, the height of the tree is 8√3 meters.

3. A contractor plans to install two slides for children to play in a park. For children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 meters 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 meters and is inclined at an angle of 60° to the ground. What should be the length of the slide in each case?

Solution:

In the figure, l1 is the length of the slide made for children below the age of 5 years, and l2 is the length of the slide made for older children.

(for children below 5 years)

In the given figure, AB = 1.5 m, AC = l1 m, and ∠ACB = 30°, PQ = 3 m, OP = l2 m, and ∠POQ = 60°.

We have:

AB / AC = sin 30°

PQ / OP = sin 60°

Substituting the given values:

1.5 / l1 = 1/2

3 / l2 = √3 / 2

Solving for l1:

l1 = 2 × 1.5 m

l1 = 3 m

Solving for l2:

l2 = (3 × 2) / √3 m

l2 = 6 / √3 m

l2 = (6√3) / 3 m

l2 = 2√3 m

Therefore, the lengths of the slides are 3 m and 2√3 m, respectively.

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 AB be the height of the tower, denoted as h. Let C be the point on the ground which is 30 m away from the foot of the tower (B). We are given that ∠ACB = 30°.

Therefore, AC = 30 m.

Now, AB/AC = tan 30°

=> h/30 = 1/√3

=> h = 30/√3 = (30/√3) * (√3/√3) = 10√3

Thus, the required height of the tower is 10√3 m.

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:

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

Let OP = l be the length of the string.

∠POQ = 60° (Given)

Now, PQ/OP = sin 60°

=> 60/l = sin 60° = √3/2

=> 60/l = √3/2

=> l = 40√3 m

Therefore, the length of the string is 40√3 m.

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:

PQ = 30 m is the height of the building.

OA = 1.5 m is the height of the boy. Its first position is at OA. OR is the horizontal line through the position of the eye at O.

∠POR = 30° (Given)

The second position of the boy is at O'A' and ∠PO'R = 60°.

Here, RQ = OA = O'A' = 1.5 m

and PR = 30 m - 1.5 m = 28.5 m.

The distance walked by the boy towards the building = R'R

[From (i) and (ii)]

= 28.5 × √3 m - 28.5/√3 m

= 28.5 × {√3 - 1/√3} m

= 28.5 × (3 - 1)/√3 m = 28.5 × 2/√3 m

= 57/√3 m = 19√3 m

Therefore, the boy walks 19√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 20 m high building are 45° and 60° respectively. Find the height of the tower.

Solution:

PQ = 20 m is the height of the building.

Let PR = h meters be the height of the transmission tower. P is the bottom and R is the top of the transmission tower.

∠POQ = 45° and ∠ROQ = 60°

In ΔOPQ,

PQ/OQ = tan 45°

=> 20/OQ = 1

=> OQ = 20 m

In triangle ORQ,

RQ/OQ = tan 60°

=> (20 + h)/20 = √3

(Since RQ = PQ + PR = 20 + h meters and OQ = 20 meters)

=> 1 + h/20 = √3

=> h/20 = (√3 - 1)

=> h = 20(√3 - 1) m

Therefore, the height of the tower is 20(√3 - 1) m.

8. A statue, 1.6 m tall, stands 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 same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.

Solution:

In the figure, DC represents the statue and BC represents the pedestal.

Let the height of the pedestal be 'h'.

Now, in right triangle ABC, we have

AB/BC = cot 45° = 1

=> AB/h = 1

=> AB = h meters

Now in right triangle ABD, we have

BD/AB = tan 60° = √3

=> BD = √3 × AB = √3 × h [from (i)]

=> h + 1.6 = √3h

=> h(√3 - 1) = 1.6

=> h = 1.6 / (√3 - 1) = 1.6 / (√3 - 1) × (√3 + 1) / (√3 + 1)

=> h = 1.6 / (3 - 1) × (√3 + 1) = 1.6 / 2 × (√3 + 1)

= 0.8(√3 + 1) m

Thus, the height of the pedestal is 0.8(√3 + 1) m.

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:

PQ = 50 meters is the height of the tower. Let AB = h meters be the height of the building. Angle of elevation of the top of the building from the foot of the tower = 30°, i.e., ∠AQB = 30°. (Top of Tower)

Angle of elevation of the top of the tower from the foot of the building = 60°, i.e., ∠PBQ = 60°.

In ΔAQB, AB/BQ = tan 30°

h/BQ = 1/√3

=> BQ = h√3  ...(i)

In ΔPBQ, 50/BQ = tan 60° = √3

=> BQ = 50/√3 ...(ii)

From (i) and (ii), we have h√3 = 50/√3

=> h = 50/3 m, i.e., h = 16 2/3 m.

Therefore, the height of the building is 16 2/3 m.

10. Two poles of equal heights are standing opposite to each other on 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 the poles and the distance of the point from the poles.

Solution:

Let AB and CD be two poles and P is the point in between them.

AB = h meters

=> CD = h meters

AP = x meters

=> CP = (80 - x) meters

Now, in right triangle APB, we have:

AB / AP = tan 60°

=> h / x = √3

=> h = √3x  ...(i)

Again, in right triangle CPD, we have:

CD / CP = tan 30°

=> h / (80 - x) = 1 / √3

=> h = (80 - x) / √3  ...(ii)

From (i) and (ii), we get:

√3x = (80 - x) / √3

=> √3 × √3 × x = 80 - x

=> 3x = 80 - x

=> 3x + x = 80

=> 4x = 80

=> x = 80 / 4 = 20

Therefore, CP = 80 - x = 80 - 20 = 60 meters.

Now, from (i), we have:

h = √3 × 20 = 1.732 × 20 = 34.64 meters.

Thus, the required point is 20 meters away from the first pole and 60 meters away from the second pole.

Height of each pole = 34.64 meters.

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 60°. From another point 20 meters 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° (see figure). Find the height of the tower and the width of the canal.

Solution:

Let PQ = h meters be the height of the tower and BQ = x meters be the width of the canal.

∠PBQ = 60°.

Now, the angle of elevation of the top of the tower from point A is 30°.

i.e., ∠PAQ = 30° where AB = 20 metres.

In ΔPBQ,

h/x = tan 60° => h/x = √3

=> h = √3x  ...(i)

In ΔPAQ,

h/(20+x) = tan 30° = 1/√3

=> h = (20+x)/√3  ...(ii)

From (i) and (ii), we have √3x = (20+x)/√3

=> 3x = 20+x

=> 2x = 20 => x = 10

From (i), h = 10√3 m

Therefore, the height of the tower is 10√3 m and the width of the canal is 10 m.

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:

Let PQ = h metres be the height of the cable tower.

AB = 7 metres is the height of the building.

∠PAR = 60° is the angle of elevation of the top of the cable tower from the top of the building.

∠RAQ = 45° is the angle of depression of the foot of the cable tower from the top of the building. Then ∠AQB = 45°.

Now, BQ = AR = x meters (say)

In triangle AQB, AB/BQ = tan 45° => 7/x = 1 => x = 7 meters.

Now, in triangle PAR, PR/AR = tan 60°

=> (PQ - QR)/x = √3

=> (h - 7)/x = √3

=> h - 7 = x√3

=> h = 7√3 + 7

=> h = 7(√3 + 1)

Hence, the height of the cable tower is 7(√3 + 1) meters.

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:

In the figure, let AB represent the lighthouse.

Therefore, AB = 75 meters.

Let the two ships be C and D such that the angles of depression from A are 45° and 30° respectively.

Now, in the right triangle ABC, we have AB/BC = tan 45°.

=> 75/BC = 1

=> BC = 75 m

Again, in right triangle ABD, we have

AB/BD = tan 30°

=> 75/BD = 1/√3

=> BD = 75√3 m

Since the distance between the two ships

= CD = BD - BC = 75√3 - 75 = 75(√3 - 1)

= 75(1.732 - 1) = 75 × 0.732 = 54.9 m

Thus, the required distance between the ships is 54.9 m.

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° (see figure). Find the distance travelled by the balloon during the interval.

Solution:

From the figure, we have ∠AOB = 60° which is the angle of elevation for the first position of the balloon. Let OB = x m.

(1st Position of balloon)

(2nd Position of balloon)

We are given that AC = 88.2 m, and AB = 88.2 - 1.2 = 87 m.

For the second position of the balloon, we have ∠POQ = 30°. Let OQ = y.

We have to find the distance d = BQ = (y - x).

We know that AB/OB = tan 60° and PQ/OQ = tan 30°.

This implies 87/x = √3 and 87/y = 1/√3.

Therefore, x = 87/√3 m and y = 87√3 m.

Now, d = y - x = (87√3 - 87/√3) m.

d = 87 × (√3 - 1/√3) m.

d = 87 × (2/√3) m.

d = 87 × (2/3) × √3 m.

d = (174/3)√3 m.

d = 58√3 m.

Therefore, the distance travelled by the balloon during the interval is 58√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 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:

Let PQ = h meters be the height of the tower. P is the top of the tower. PX is a horizontal line through P. The first and second positions of the car are at A and B respectively.

∠APX = 30°

Angles of depression of the car when observed at A and B are ∠APX = 30° and ∠BPX = 60° respectively.

Then, ∠PAQ = 30° and ∠PBQ = 60°

Let the speed of the car be x m/sec.

Then, distance AB = 6x m.

Let the time taken from B to Q be n seconds.

Then distance BQ = nx m.

In right triangle PAQ:

h / (6x + nx) = tan 30° = 1/√3

=> h = (n + 6)x / √3 ... (i)

In right triangle PBQ:

h / nx = tan 60° = √3

=> h = nx√3 ... (ii)

From (i) and (ii), we have:

(n + 6)x / √3 = nx√3

=> n + 6 = n√3 × √3

=> 3n = n + 6

=> 2n = 6

=> n = 3

Hence, the time taken to travel from B to Q = 3 seconds.

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 the tower be represented by AB in the figure.

Let AB = h meters.

Therefore, in the right triangle ABC, we have AB/AC = tan θ.

This implies h/9 = tan θ  (Equation i)

In the right triangle ABD, we have AB/AD = tan(90° - θ) = cot θ.

This implies h/4 = cot θ  (Equation ii)

Multiplying equations (i) and (ii), we get:

(h/9) x (h/4) = tan θ x cot θ = 1  (Since tan θ x cot θ = 1)

This simplifies to h²/36 = 1.

Therefore, h² = 36.

Taking the square root, h = ± 6 meters.

Since height is positive only, h = 6 meters.

Thus, the height of the tower is 6 meters.

5.0Benefits of Solving NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry

• Helps to apply trigonometry to real-world scenarios like measuring the height of buildings or the distance between objects.
• Provides step-by-step explanations, making it easier to grasp trigonometric concepts and applications and thus strengthening conceptual clarity
• Encourages logical thinking by solving practical problems using trigonometric ratios.
• Improves accuracy in applying trigonometric formulas and solving complex numerical problems