MCQ: Height & Distance - 1 - SSC CGL MCQ

# MCQ: Height & Distance - 1 - SSC CGL MCQ

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## 15 Questions MCQ Test Quantitative Aptitude for SSC CGL - MCQ: Height & Distance - 1

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

### The angle of elevation of the sun is 60°. Find the length of the shadow of a man who is 180 cm tall.

Detailed Solution for MCQ: Height & Distance - 1 - Question 1

Let AB be the man and BC be his shadow

⇒ AB = 180

⇒ tan60° = AB/BC

√3 = 180/BC

∴ BC = 103.92 cm

MCQ: Height & Distance - 1 - Question 2

### The stature of a tree is 10m. It is twisted by the wind in a manner that its top touches the ground and makes a point of 60 with the ground. At what range from base did the tree get twisted? (√3 = 1.73)

Detailed Solution for MCQ: Height & Distance - 1 - Question 2

Let AB be the tree bowed at the point C so that part CB takes the position CD.
Then, CD = CB.
Let AC = x meters. At that point, CD = CB = (10 - X) m and ∠ADC = 60°.
AC/AC = sin60° => x/(10 - x) = √3/2
=> 2x = 10 √3 - √3x
=> (2 + √3) x = 10 √3
=>x = 10 √3/ (2 + √3) * (2 - √3)/(2 - √3) = 20 √3 - 30)m
= (20 * 1.73 - 30) m = 4.6m
=> Required height = 4.6m.

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MCQ: Height & Distance - 1 - Question 3

### What is the distance between a tower and an observer if the angle of elevation from the observer’s eye to the top of the tower (height = 50 m) is 30°? The observer is 1.5 m tall.

Detailed Solution for MCQ: Height & Distance - 1 - Question 3

Let AB be the tower and CD be the observer

tan30° = AE/CE

⇒ 1/√3= (AB – CD) / x = (50 – 1.5) / x = 48.5/x

x = 84 m

MCQ: Height & Distance - 1 - Question 4

A kite is flying at a tallness of 75 m from the level of ground, joined to a string slanted at 60° to the level. The string's length is:

Detailed Solution for MCQ: Height & Distance - 1 - Question 4

Let AB be the kite and AC be the level ground
So that BC - AC.
At that point, ∠BAC = 60°and BC = 75m. Let AB = x meters.
Presently AB/BC = coses60° = 2/ √3
=> x/75 = 2/√3 => x = 150/√3 = 150 * √3/3 = 50 √3m.
∴ Length of the string = 50 √3m.

MCQ: Height & Distance - 1 - Question 5

Two boats are spotted on the two sides of a light house. If the angle of depression made by both the boats from top of the lighthouse is 30° and 45° and the height of the light house is 125 m then find the distance between the two boats.

Detailed Solution for MCQ: Height & Distance - 1 - Question 5

Let AB be the light house and the two boats be at C and D

∴ AB = 125 m

tan30° = BC/AB = x/125 = 1/3

=> x = 72.17 m

tan45° = BD/AB = y/125 = 1

=> y = 125 m

∴ The distance between the two boats is = x + y

= 72.17 + 125

= 197.17 m

MCQ: Height & Distance - 1 - Question 6

At a moment, the shadow's length of a shaft is √3 times the stature of the shaft. The edge of rise of the sun is:

Detailed Solution for MCQ: Height & Distance - 1 - Question 6

Let AB be the post and AC be its shadow.
Let AB = X m. Then, AC = √3Xm.Let∠ACB = θ.
AB/AC = tanθ => tanθ = X/√3X = 1/√3 = tan30°.
∴ θ = 30°.
Hence, the point of rise is 30°.

MCQ: Height & Distance - 1 - Question 7

A man standing on the terrace of a building watches a car speeding towards him. If at that particular instant the car is 200 m away from the building makes an angle of depression of 60° with the man’s eye and after 8 seconds the angle of depression is 30°, what is the speed of the car?

Detailed Solution for MCQ: Height & Distance - 1 - Question 7

Let AB be the building and the man is standing at A.

When the car is 200m away from the building, the angle of depression is 60°

tan 60° = BD/AB = √3

⇒ 200/AB = √3

⇒ AB = 115.47 m

⇒ tan30° =BC/AB = x/115.47 = 1/√3

⇒ x = 66.67

Now, the car travels distance CD in 8 seconds

CD = BD – BC = 200 – 66.67 = 133.33 m

Speed = 133.33/8 = 16.67 m/s

MCQ: Height & Distance - 1 - Question 8

Two men are inverse sides of a tower. They gauge the edge of the rise of the highest point of the tower as 30° and 45° respectively. On the off chance that the tallness of the tower is 50 m, discover the separation between the two men. (Take √3=1.732)

Detailed Solution for MCQ: Height & Distance - 1 - Question 8

Let AB be the tower and let C and D be the two's positions men.
At that point ∠ACB = 30°,∠ADB = 45°and AB = 50 m
AC/AB = Cot30° = √3 => AC/50 = √3
=> AC = 50√3m
Separation between the two men = CD = (AC + AD)
= (50√3 + 50) m = 50(√3 + 1)
=50(1.73 + 1)m = (50 * 2.73)m = 136.5m.

MCQ: Height & Distance - 1 - Question 9

A ship is approaching an observation tower. If the time taken by the ship to change the angle of elevation from 30° to 45° is 10 minutes, then find the time the ship will take to cover the remaining distance and reach the observation tower assuming the ship to be travelling at a uniform speed.

Detailed Solution for MCQ: Height & Distance - 1 - Question 9

Let AB be the observation tower and h be its height.

Also, let the ship be at C when the angle of elevation is 30° and at D when the angle of elevation is 45°.

The time taken by the ship to travel from C to D is 10 minutes and we need to find out the time the ship will take to reach B from D.

∴ tan 30º = AB/CB = h / CB = 1/ √3

⇒ CB = √3 x h

∴ tan 45º = AB / DB = h / DB = 1

⇒ DB = h

⇒ CD = CB – DB = (√3h – h) = h(√3– 1)

Now, as h(√3 – 1) distance is covered in 10 minutes, a distance of h is covered in = 13.66 minutes = 13 minutes 40 seconds

MCQ: Height & Distance - 1 - Question 10

The shadow of a building is 10 m long when the point of rise of the sun is 60°. Discover the building's stature.

Detailed Solution for MCQ: Height & Distance - 1 - Question 10

Let AB be the building and AC be its shadow.
Then, AC = 20m and ∠ACB = 60°. Let AB = x m.
Presently AB/AC = tan 60° = √3 => x/10 = √3
=> x = 10√3m = (10 * 1.732) m = 17.32m.
∴ Height of the building is 17.32m.

MCQ: Height & Distance - 1 - Question 11

On the two sides of a road are two tall buildings exactly opposite to each other. The height of the taller building is 60 m. If the angle of elevation from the top of the smaller building to the top of the taller one is 30° and the angle of depression from top of the taller building to the foot of the smaller one is 30°, then find the height of the smaller building.

Detailed Solution for MCQ: Height & Distance - 1 - Question 11

Let AB be the taller building of height 60 m and CD be the smaller one of height h m.

⇒ DB / AB = tan 30º = 1/√3

⇒ DB = AB x tan 30º

= 34.64 m.

tan 30º = AE/CE = AE/DB = 1/ = AE/34.64

∴ AE = 60 – h = 20

∴ h = 40 m

MCQ: Height & Distance - 1 - Question 12

A straight tree is broken due to thunder storm. The broken part is bent in such a way that the peak touches the ground at an angle elevation of 45°. The distance of peak of tree (where it touches the root of the tree is 20 m. Then the height of the tree is

Detailed Solution for MCQ: Height & Distance - 1 - Question 12

Let the total length of the tree be X + Y meters
From the figure tan 45 = X/20 => X = 20
cos 45 = 20/Y => Y = 20/cos 45 = 20√2
X + Y = 20 + 20radic; 2 = 20 + 2 x 10 x 1.414 = 48.28 meters

MCQ: Height & Distance - 1 - Question 13

An observer 1.6 m tall is 20√3 away from a tower. The angle of elevation from his eye to the top of the tower is 30º. The heights of the tower is:

Detailed Solution for MCQ: Height & Distance - 1 - Question 13

Let AB be the observer and CD be the tower.

Draw BE ⊥ CD.

Then, CE = AB = 1.6 m,

BE = AC = 20√3 m.

∴ CD = CE + DE = (1.6 + 20) m = 21.6 m.

MCQ: Height & Distance - 1 - Question 14

An aeroplane when 750 m high passes vertically above another aeroplane at an instant when their angles of elevation at same observing point are 45° and 30&de; respectively. Approximately, how many meters higher is the one than the other?

Detailed Solution for MCQ: Height & Distance - 1 - Question 14

Let C and D be the position of the aeroplanes.
Given that CB = 900 m,∠CAB = 60°,∠DAB = 45°
From the right △ ABC,
Tan45 = CB/AB => CB = AB
Tan30 = DB/AB => DB = ABtan30 = CBx(1/√3) = 750/√3
CB = CD + DB
=> Required height CD = CB - DB = 750 - 750/√3 = 250(3 - √3)

MCQ: Height & Distance - 1 - Question 15

The angle of elevation of a ladder leaning against a wall is 60º and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is:

Detailed Solution for MCQ: Height & Distance - 1 - Question 15

Let AB be the wall and BC be the ladder.

Then, ACB = 60º and AC = 4.6 m.

⇒ BC = 2 x AC
= (2 x 4.6) m
= 9.2 m.

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