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