MCQ Test: Heights and Distances - 2 - SSC CGL MCQ

# MCQ Test: Heights and Distances - 2 - SSC CGL MCQ

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## 15 Questions MCQ Test Quantitative Aptitude for SSC CGL - MCQ Test: Heights and Distances - 2

MCQ Test: Heights and Distances - 2 for SSC CGL 2024 is part of Quantitative Aptitude for SSC CGL preparation. The MCQ Test: Heights and Distances - 2 questions and answers have been prepared according to the SSC CGL exam syllabus.The MCQ Test: Heights and Distances - 2 MCQs are made for SSC CGL 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for MCQ Test: Heights and Distances - 2 below.
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MCQ Test: Heights and Distances - 2 - Question 1

### At a point on a horizontal line through the base of a monument, the angle of elevation of the top of the monument is found to be such that its tangent is 1/5. On walking 138 metres towards the monument the secant of the angle of elevation is found to be √(193) / 12. The height of the monument (in meter) is

Detailed Solution for MCQ Test: Heights and Distances - 2 - Question 1

Given, the distance walking, CD = 138 m
Let, The height of the monument, AB = h metre
BD = x metre, ∠ACB = α and ∠ADB = β

∴ BC = CD + BD = (138 + x) m
We know that,

x = 5h – 138 ...(i)
Now, in ΔABD,

⇒ 7x = 12h
⇒ 7(5h – 138) = 12h [From eq. (i)]
⇒ 35h – 966 = 12h ⇒ 23h = 966 ⇒ h = 42 m
∴ The height of the monument is 42 metre.
Hence, option C is correct.

MCQ Test: Heights and Distances - 2 - Question 2

### The angle of elevation of the top of a tower from two points A and B lying on the horizontal through the foot of the tower are respectively 15° and 30°. If A and B are on the same side of the tower and AB = 48 metre, then the height of the tower is :

Detailed Solution for MCQ Test: Heights and Distances - 2 - Question 2

Given, AB = 48 m
Let, the height of the tower, CD = h metre
And, BC = x metre
∴ AC = AB + BC = (48 + x) m
In ΔBCD,

x = h √3 ...(i)
Now, in ACD,

⇒ h = 96 + 2h √3 – 48 √3 – 3h
⇒ 4h – 2h √3 = 48(2 – √3)
⇒ 2h(2 –√3) = 48(2 – √3)
⇒ h = 24 m
∴ The height of the tower is 24 metre.
Hence, option B is correct.

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MCQ Test: Heights and Distances - 2 - Question 3

### If the angle of elevation of the sun changes from 30° to 45°, the length of the shadow of a pillar decreases by 20 metres. The height of the pillar is

Detailed Solution for MCQ Test: Heights and Distances - 2 - Question 3

Let, the height of the pillar, AB = h metre.
When the sun's angle of elevation was 30°, then the length of shadow of the pillar is BD.
And, when the sun's angle of elevation is 45°, then the length of shadow of the pillar is BC = x metre (let).
When the sun changes from 30° to 45°, then the length of shadow of the pillar decreases CD = 20 (given)
∴ BD = BC + CD = (x + 20) m
In ΔABC,

⇒ h = x ...(i)
Now, in ΔABD,

⇒ h 3 = x + 20
⇒ h √3 = h + 20 [From eq. (i)]
⇒ h ( √3 – 1) = 20

∴ The height of the pillar is 10 (√3 + 1) metre.
Hence, option D is correct.

MCQ Test: Heights and Distances - 2 - Question 4

From a point P on the ground the angles of elevation of the top of a 10 m tall building is 30°. A flag is hoisted at the top of the building and the angle of elevation of the top of the flagstaff from P is 45° Find the length of the flagstaff. (Take √3 = 1.732)

Detailed Solution for MCQ Test: Heights and Distances - 2 - Question 4

AC = Flag, AB = building = 10 m
∠APB = 30°; ∠CPB = 45°
In Δ APB,

⇒ PB = 10 √3 m
In ΔPBC,

⇒ PB = AB + AC ⇒ 10 √3 = 10 + AC
⇒ AC = 10 √3 – 10
⇒ 10(√3 – 1) m = 10(1.732 – 1) m
= 10 × 0.732 = 7.32 m.
Hence, option D is correct.

MCQ Test: Heights and Distances - 2 - Question 5

The top of two poles of height 24 m and 36 m are connected by a wire. If the wire makes an angle of 60° with the horizontal, then the length of the wire is

Detailed Solution for MCQ Test: Heights and Distances - 2 - Question 5

DE = 36 – 24 = 12 m

Hence, option B is correct.

MCQ Test: Heights and Distances - 2 - Question 6

The angle of elevation of an aeroplane from a point on the ground is 60°. After 15 seconds flight, the elevation changes to 30°. If the aeroplane is flying at a height of 1500√3 m, find the speed of the plane.

Detailed Solution for MCQ Test: Heights and Distances - 2 - Question 6

Given, the height of the aeroplane, BC = 1500 √3 meter
Let, an aeroplane travelled distance BD = x metre in 15 seconds.
And, AC = y metre
∴ AE = CE + AC = (x + y) m
In ΔABC,

y = 1500 m

x + y = 4500
x + 1500 = 4500 x = 3000 m
∴ The aeroplane travelled 3000 m distance in 15 seconds.
∴ Speed = 3000/15 = 200 m/sec
Hence, option A is correct.

MCQ Test: Heights and Distances - 2 - Question 7

The shadow of a tower is √3 times its height. Then the angle of elevation of the top of the tower is

Detailed Solution for MCQ Test: Heights and Distances - 2 - Question 7

Let, the height of the tower, AB = h metre
And, the angle of elevation = Θ
then, the shadow of the tower, BC = h√3 metre
In ΔABC,

Θ = 30°
Hence, option B is correct.

MCQ Test: Heights and Distances - 2 - Question 8

A vertical post 15 ft high is broken at a certain height and its upper part, not completely separated, meets the ground at an angle of 30°. Find the height at which the post is broken.

Detailed Solution for MCQ Test: Heights and Distances - 2 - Question 8

Total height of post = AB + AC = 15 ft
Let, the post is broken at point A, ∴ AB = h
And, AC = 15 – AB = (15 – h) ft
In ΔABC,

15 – h = 2h
3h = 15
h = 5 ft
Hence, option B is correct.

MCQ Test: Heights and Distances - 2 - Question 9

The shadow of the tower becomes 60 metres longer when the altitude of the sun changes from 45° to 30°. Then the height of the tower is

Detailed Solution for MCQ Test: Heights and Distances - 2 - Question 9

Let, the height of the tower, AB = h metre.
When the sun's angle of elevation was 45°, then the length of shadow of the tower is BD = x (let).
When the sun changes from 45° to 30°, then the length of shadow of the tower increases CD = 60 m (given)
And, when the sun's angle of elevation is 30°, then the length of shadow of the tower is BC = CD + BD = (60 + x) metre.
In ΔABD,

h = x
Now, in ΔABC,

h√3 – x = 60
h√3 – h = 60 [∵ h = x]
h(√3 – 1) = 60

= 30(√3 + 1) m
Hence, option C is correct.

MCQ Test: Heights and Distances - 2 - Question 10

An aeroplane when flying at a height of 3125 m from the ground passes vertically below another plane at an instant when the angle of elevation of the two planes from the same point on the ground are 30° and 60° respectively. The distance between the two planes at that instant is

Detailed Solution for MCQ Test: Heights and Distances - 2 - Question 10

Given, the height of an aeroplane from the ground, BD = 3125 m
Let, the distance between the aeroplanes, AD = x metre
And, BC = y metre
∴ AB = AD + BD = (3125 + x) m
In ΔBCD,

y√3 = x + 3125
3125√3 × √3 = x + 3125
x = 9375 – 3125 = 6250 m
∴ The distance between the aeroplanes is 6250 metres.
Hence, option D is correct.

MCQ Test: Heights and Distances - 2 - Question 11

The distance between two vertical poles is 60 m. The height of one of the poles is double the height of the other. The angle of elevation of the top of the poles from the middle point of the line segment joining their feet are complementary to each other. The height of the poles are:

Detailed Solution for MCQ Test: Heights and Distances - 2 - Question 11

Given, BC = 60 metre,
Let E is the mid-point of BC
∴ BE = EC = 30 metre
Let, the height of the pole AB = h metre
∴ The height of the pole CD = 2h metre
And, ∠AEB and ∠DEC are complementary.
∴ ∠AEB = (90° – Θ) and ∠DEC = Θ
In ΔABE,

Now, in ΔCDE,

By multiplying both equation (i) and (ii),

∴ The height of the pole AB = h = 15 √2
And, the height of the pole CD = 2h = 2 × 15 √2 = 30 √2
Hence, option D is correct.

MCQ Test: Heights and Distances - 2 - Question 12

A man standing at a point P is watching the top of a tower, which makes an angle of elevation of 30°. The man walks some distance towards the tower and then his angle of elevation of the top of the tower is 60°. If the height of the tower is 30 m, then the distance he moves is

Detailed Solution for MCQ Test: Heights and Distances - 2 - Question 12

Given, the height of the tower, AB = 30 m
Let, the man moves the distance, CP = y metre
And, BC = x metre
∴ BP = BC + CP = (x + y) m
In ΔABC,

Hence, option A is correct.

MCQ Test: Heights and Distances - 2 - Question 13

An aeroplane when flying at a height of 5000 m from the ground passes vertically above another aeroplane at an instant, when the angles of elevation of the two aeroplanes from the same point on the ground are 60° and 45° respectively. The vertical distance between the aeroplanes at that instant is

Detailed Solution for MCQ Test: Heights and Distances - 2 - Question 13

Given, the height of an aeroplane from the ground, AB = 5000 m,
Let, the distance between the aeroplanes, AD = x metre
And, BC = y metre
∴ BD = AB – AD = (5000 – x) m
In ΔABC,

Hence, option C is correct.

MCQ Test: Heights and Distances - 2 - Question 14

Two posts are x metres apart and the height of one is double that of the other. If from the mid-point of the line joining their feet, an observer finds the angular elevations of their tops to be complementary, then the height (in metres) of the shorter post is

Detailed Solution for MCQ Test: Heights and Distances - 2 - Question 14

Given, BC = x metre,
Let E is the mid-point of BC
∴ BE = EC = x/2 metre
Let, the height of the pole AB = h metre
∴ The height of the pole CD = 2h metre
And, ∠AEB and ∠DEC are complementary.
∴ ∠AEB = (90° – Θ) and ∠DEC = Θ
In ΔABE,

[∵ tan (90° – Θ) = cot Θ]
Now, in ΔCDE,

By multiplying both equation (i) and (ii),

Hence, option A is correct.

MCQ Test: Heights and Distances - 2 - Question 15

The angle of elevation of the top of a building and the top of the chimney on the roof of the building from a point on the ground are x° and 45° respectively. The height of building is h metre. Then the height of the chimney (in metre) is :

Detailed Solution for MCQ Test: Heights and Distances - 2 - Question 15

Given, the height of the building, BD = h metre
Let, the height of the chimney, AD = x metre
and, BC = a metre
∴ AB = BD + AD = (h + x) m
In ΔABC,

a = h cot x ...(ii)
From eq. (i) and (ii),
x + h = h cot x
x = h cot x – h
Hence, option B is correct.

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## Quantitative Aptitude for SSC CGL

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