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

Download Free Trignometry PDFs | Mathematics Quiz Questions And Answers for SSC CHSL 2020, SSC CGL TIER I and SSC CGL TIER II

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,

tanΘ = AB / BQ

tan Θ = h / b .....(i)

Now, in ΔABP,

tan (90º – Θ) = AB / BP

cot Θ = (h / a) .....(ii)

[∵  tan (90° – Θ) = cot Θ]

By multiplying both equations (i) and (ii)

tan Θ cot Θ = (h / b) × (h / a)

1 = (h2 / ab) [∴ tan Θ cot Θ = 1]

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 30º = AB / BC

(1 / √3) = (h / 100)

h = (100 / √3) m

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,

tab Θ = AB / BC = h / 100 .....(i)

Now, in ΔABD,

tan 2Θ = (AB / BD)

2tanΘ / (1 – tan2Θ) = h / 60

[∵ tan 2 Θ = 2tanΘ / (1 – tan2 Θ)]

120 tan Θ = h(1 – tan2 Θ)

120 × (h / 160) = h {1 – (h / 160)2}

[From eq..... (i)]

3 / 4 = 1 – (h / 160)2

(h / 160)2 = 1 – (3 / 4) = 1 / 4

h / 160 = √(1 / 4) = 1 / 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 30º = AB / BC

1 / √3 = h / 50

h = 50 / √3 m
 
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,

tan45º = AB / BD ⇒ 1 = h / x
 
⇒  h = x ...(i)

Now, in ΔABC,

tan 30º = AB / BC ⇒ 1 / √3

= h / (x + 10)

⇒ h√3 – x = 10

⇒ h√3 – h = 10     [From eq. (i)]

⇒ h(√3 – 1) = 10

⇒ h = 10 / (√3 – 1) × {(√3 + 1) / (√3 + 1)}

= 5 (√3 + 1)m

Hence, option A is correct.