NCERT Exemplar: Some Applications of Trigonometry

Q.1. If a pole 6 m high casts a shadow 2√3 m long on the ground, find the Sun's elevation.
Ans: 

tan θ = opposite/adjacent = 6/(2√3)
= 6/(2√3) = 3/√3 = √3
∴ θ = 60°

Q.2. If the angle of elevation of a tower from a distance of 100 m from its foot is 60°, then what will be the height of the tower?

Ans: Let h be the height of the tower.

tan 60° = h/100
⇒ √3 = h/100
⇒ h = 100√3 m

Q.3. A spherical balloon of radius r subtends an angle θ at the eye of an observer. If the angle of elevation of its centre is φ, find the height of the centre of the balloon.
Ans: In Fig. 11.55, O is the centre of the balloon, whose radius OP = r and ∠PAQ = θ. Also, ∠OAB = φ. Let the height of the centre of the balloon be h. Thus OB = h.
In ΔOAP, we have

Also in ΔOAB,
 ...(ii)
From (i) and (ii), we get
 or 

Q.4. If the angle of elevation of a cloud from a point h metres above a lake is α and the angle of depression of its reflection in the lake is β, prove that the height of the cloud is

Ans: Let AB be the surface of the lake and let P be a point of observation (Fig. 11.58) such that AP = h metres. Let C be the position of the cloud and C' be its reflection in the lake. Then, CB = C'B. Let PM be perpendicular from P on CB. Then, ∠CPM = α and ∠MPC' = β. Let CM = x.
Then, CB = CM + MB = CM + PA = x + h
In ΔCPM, we have
 

⇒ AB = x cot α
In ΔPMC', we have
 

⇒ AB = (x + 2h) cotβ ...(ii)
From (i) and (ii), we have


Hence, the height CB of the cloud is given by
CB = x + h ⇒
⇒ 

Q.5. The lower window of a house is at a height of 2 m above the ground and its upper window is 4 m vertically above the lower window. At a certain instant, the angles of elevation of a balloon from these windows are observed to be 60° and 30° respectively. Find the height of the balloon above the ground.
Ans: Let the height of balloon above ground be h m and let the horizontal distance from the windows to the balloon be x m.
Lower window is at 2 m above ground and upper window is at 2 + 4 = 6 m above ground.
In right-angled triangle formed with the lower window,

tan 60° = (h - 2)/x = √3 ⇒ h - 2 = x√3 ...(i)
 In right-angled triangle formed with the upper window,

tan 30° = (h - 6)/x = 1/√3 ⇒ h - 6 = x/√3 ...(ii)
Subtract (ii) from (i):
(h - 2) - (h - 6) = x√3 - x/√3
⇒ 4 = x(√3 - 1/√3) = x(2/√3)
⇒ x = 4 × (√3/2) = 2√3 m
Use (i): h - 2 = x√3 = 2√3 × √3 = 2 × 3 = 6
⇒ h = 8 m


Height of the balloon above the ground is 8 m.