NCERT Solutions: Some Applications of Trigonometry (Exercise 9.2)

NCERT Solutions for Class 10 Maths Chapter 9 - Some Applications of Trigonometry (Exercise 9.2)

Exercise 9.2

Q1. The angles of depression of the top and the bottom of a building 50 m high as observed from the top of a tower are 30° and 60° respectively. Find the height of the tower and also the horizontal distance between the building and the tower.

Ans: In the figure

Let AB = 50 m be the building.
Let CE be the tower such that CE = (50 + x) m
In right ΔADE, we have:

⇒         ...(1)

In right ΔACE, we have:

⇒      ...(2)

From (1) and (2), we get

⇒
⇒ 3x − x =50 ⇒ x = 25

∴ Height of the tower = 50 + x
= 50 + 25
= 75 m

Now from (1), BC = √3 × x
= √3 × 25 m
= 1.732 × 25 m
= 43.25 m

i.e., The horizontal distance between the building and the tower = 43.25 m.

Q2. The angle of elevation of the top of a tower as observed from a point on the ground is ‘α ’ and on moving ‘a’ metres towards the tower, the angle of elevation is ‘β’. Prove that the height of the tower is

Ans: In the figure, let the tower be represented by AB.
∴ In right Δ ABC, we have:

⇒ x tan β = h

⇒

Now, in right ΔABD, we have:

Q3. A vertical tower stands on a horizontal plane and is surmounted by a vertical flag staff of height 5 m. From a point on the plane the angles of elevation of the bottom and top of the flag staff are respectively 30° and 60°. Find the height of the tower.

Ans: Let in the figure, BC be the tower such that

BC = y metres.

CD be the flag staff such that
CD = 5m
⇒ BD =(y + 5) m.

In right Δ ABC, we have:

⇒      ...(1)

In right Δ ABD, we have:

⇒

∴

⇒ y + 5 = 3 y
⇒ 3y − y = 5 ⇒ y = 5/2  = 2.5 m

∴ The height of the tower = 2.5 m.

Q4. The length of the shadow of a tower standing on level plane is found to be 20 m longer when the sun’s altitude is 30° than when it was 60°. Find the height of the tower.

Ans: In the figure, let CD be the tower such that
CD = h metres
Also BC = x metres

In right Δ ACD, we have:

⇒

⇒

Thus, the height of the tower = 17.32 m.

Q5. From the top of a hill 200 m high, the angles of depression of the top and bottom of a pillar are 30° and 60° respectively. Find the height of the pillar and its distance from the hill.

Ans: In the figure, let AD is the hill such that
AD = 200 m and CE is the pillar.

∴

⇒

⇒ Distance between pillar and hill = 115.33 m

Now,      [∵ DE = BC]

In right ∆ ABC, we have:

⇒

∴ Height of the pillar

CE = AD − AB    [∴ CE = BD]
= 200 − 66.67 m
= 133.33 m

Q6. The angles of elevation of the top of a tower from two points on the ground at distances a and b units from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is  units.

Ans: In the figure, AB is the tower, such that:

AB = h
BD = b
BC = a

In right Δ ABD, we have

⇒
⇒  h = b cot θ     ...(1)

In right Δ ABC, we have

⇒      ..(2)

Multiplying (1) and (2), we get
h × h = b cot θ × a tan θ
⇒ h2 = a × b × (cot θ × tan θ)    [∵ cot θ × tan θ = 1]
⇒ h2 = a b

Q7. A pole 5 m high is fixed on the top of a tower. The angle of elevation of the top of the pole observed from a point ‘A’ on the ground is 60° and the angle of depression of the point ‘A’ from the top of the tower is 45°. Find the height of the tower.

Ans: In the figure, let BC be the tower and CD be the pole.
Let BC = x metres and AB = y metres In right ∆ ABC, we get

⇒ BC = AB ⇒ y = x ... (1)

In right Δ ABD, we have:

⇒
⇒
⇒
∴        [∵ x = y from (1)]
⇒
⇒

Thus, the height of the tower = 6.83 m

The document NCERT Solutions for Class 10 Maths Chapter 9 - Some Applications of Trigonometry (Exercise 9.2) is a part of the Bank Exams Course NCERT Mathematics for Competitive Exams.
All you need of Bank Exams at this link: Bank Exams

NCERT Mathematics for Competitive Exams

276 docs|149 tests

FAQs on NCERT Solutions for Class 10 Maths Chapter 9 - Some Applications of Trigonometry (Exercise 9.2)

 1. What are some real-life applications of trigonometry?
Ans. Trigonometry has various real-life applications, including calculating distances, heights, and angles in navigation, architecture, engineering, and astronomy. It is also used in music to study waveforms and harmonics.
 2. How can trigonometry be used to solve problems involving angles and distances?
Ans. Trigonometry uses the relationships between angles and sides of triangles to solve problems involving angles and distances. By using trigonometric functions such as sine, cosine, and tangent, we can calculate unknown angles or distances using known values.
 3. What is the Pythagorean theorem and how is it related to trigonometry?
Ans. The Pythagorean theorem states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. Trigonometry is closely related to the Pythagorean theorem as it uses the ratios of the sides of right-angled triangles to calculate angles and distances.
 4. How is trigonometry used in architecture and engineering?
Ans. Trigonometry is used in architecture and engineering to calculate angles and distances in various structures. It helps in designing and constructing buildings, bridges, and other structures by ensuring accurate measurements and proper alignment.
 5. Can trigonometry be used in everyday life?
Ans. Yes, trigonometry can be used in everyday life. It is used in activities such as sports, navigation, and construction. For example, determining the distance between two points using a map or finding the angle of elevation for shooting a basketball are everyday applications of trigonometry.

NCERT Mathematics for Competitive Exams

276 docs|149 tests

Up next

 Explore Courses for Bank Exams exam
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;