Home » Quantitative Aptitude » Geometry » Height and Distance Level 1 » all » Height and Dist Quiz 4

Directions: Study the following questions carefully and choose the right answer:
1
The angle of elevation of the top of a tower from the point P and Q at distance of 'a' and 'b' respectively from the base of the tower and in the same straight line with it are complementary. The height of the tower is
» Explain it
A

Let, the height of the tower = h metre

Given, BP = a metre    and    BQ = b metre

And, ∠APB and ∠AQB are complementary.

∴  ∠AQB = Θ    and    ∠APB = (90° – Θ)

In ΔABQ,

tan Θ =  AB
BQ

tan Θ =  h       ...(i)
b

Now, in ΔABP,

tan (90° – Θ) =  AB
BP

cot Θ =  h       ...(ii)       [∵  tan (90° – Θ) = cot Θ]
a

By multiplying both equations (i) and (ii)

tan Θ cot Θ =  h  ×  h
b a

1 =  h2         [∵  tan Θ cot Θ = 1]
ab

h =  ab

Hence, option A is correct.

2
The angle of elevation of a tower from a distance 100 m from its foot is 30°. Height of the tower is :
» Explain it
A

Given, distance BC = 100 m

Let, the height of the tower = h metre

In ΔABC,

tan 30° =  AB
BC

1  =  h
3 100

h =  100  m
3


Hence, option A is correct.

3
A tower standing on a horizontal plane subtends a certain angle at a point 160 m apart from the foot of the tower. On advancing 100 m towards it, the tower is found to subtend an angle twice as before. The height of the tower is
» Explain it
A

Given, BC = 160 m    and    CD = 100 m

∴  BD = BC – CD = 160 – 100 = 60 m

Let, the height of the tower, AB = h metre

And, ∠ACB = Θ    and    ∠ADB = 2Θ

In ΔABC,

tan Θ =  AB  =  h       ...(i)
BC 100

Now, in ΔABD,

tan 2Θ =  AB
BD

2tanΘ  =  h     [ ∵  tan 2Θ =  2tanΘ ]
1 – tan2 Θ 60 1 – tan2 Θ

120 tan Θ = h(1 – tan2 Θ)

120 ×  h  =  h [ 1 –  ( h ) 2 ]        [From eq. (i)]
160 160  

3  = 1 –  ( h ) 2
4 160  

( h ) 2  = 1 –  3  =  1
160   4 4

h  = 
1
4
 =  1
160 2

h = 80 m

Hence, option A is correct.

4
The angle of elevation of a tower from a distance 50 m from its foot is 30°. The height of the tower is
» Explain it
B

Given, distance BC = 50 m

Let, the height of the tower AB = h metre

In ΔABC,

tan 30° =  AB
BC

1  =  h
3 50

h =  50  m
3
 
Hence, option B is correct.

5
The length of the shadow of a vertical tower on level ground increases by 10 metres when the altitude of the sun changes from 45° to 30°. Then the height of the tower is
» Explain it
A

Let, the height of the pillar, AB = h metre.
 
When the sun's angle of elevation was 45°, then the length of shadow of the pillar is BD = x (let).
 
And, when the sun's angle of elevation is 30°, then the length of shadow of the pillar is BC.
 
When the sun changes from 45° to 30°, then the length of shadow of the pillar increases CD = 10 (given)

∴  BC = CD + BD = (10 + x) m
 
In ΔABD,

tan 45° =  AB     ⇒    
1 = h
x
BD

⇒  h = x ...(i)

Now, in ΔABC,

tan 30° =  AB     ⇒    
1  =  h
3 x + 10
BC

⇒  h 3  – x = 10

⇒  h 3  – h = 10       [From eq. (i)]

⇒  h( 3  – 1) = 10

⇒  h =  10  × 
3  + 1
 = 5(
3
 + 1) m
3  – 1
3  + 1

Hence, option A is correct.