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SSC CGL Exam  >  SSC CGL Test  >  Quantitative Aptitude  >  MCQ: Height & Distance - 2 - SSC CGL MCQ

Height & Distance - 2 - Free MCQ Practice Test with solutions, SSC CGL


MCQ Practice Test & Solutions: MCQ: Height & Distance - 2 (15 Questions)

You can prepare effectively for SSC CGL Quantitative Aptitude for SSC CGL with this dedicated MCQ Practice Test (available with solutions) on the important topic of "MCQ: Height & Distance - 2". These 15 questions have been designed by the experts with the latest curriculum of SSC CGL 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 15 minutes
  • - Number of Questions: 15

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MCQ: Height & Distance - 2 - Question 1
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A man on the top of a vertical observation tower observes a car moving at a uniform speed coming directly towards it. If it takes 10 minutes for the angle of depression to change from 45° to 60°, how soon after this will the car reach the observation tower?

Detailed Solution: Question 1


Let AB be the tower and C and D be the two positions of the car.
Then, from figure
AB/AC = tan 60 = √3 => AB = √3AC
AB/AD = tan 45 = 1 => AB = AD
AB = AC + CD
CD = AB - AC = √3AC - AC = AC (√3 - 1)
CD = AC (√3 - 1) => 10 min
 AC => ?
AC/(AC(√3 - 1)) x 10 =? = 10/(√3 - 1) = 13.66 = 13 min 20 sec(approx)

MCQ: Height & Distance - 2 - Question 2
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Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30° and 45° respectively. If the lighthouse is 100 m high, the distance between the two ships is:

Detailed Solution: Question 2

Let AB be the lighthouse and C and D be the positions of the ships.

Then, AB = 100 m, ACB = 30° and ADB = 45°.


      AC = AB x 3 = 1003 m.

 CD = (AC + AD)= (1003 + 100) m
= 100(3 + 1)
= (100 x 2.73) m
= 273 m.

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MCQ: Height & Distance - 2 - Question 3
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From The highest point of a 10 m high building, the edge of rise of the of the highest point of a tower is 60° and the despondency's edge of its foot is 45°, Find The tower's stature. (take√3 = 1.732)

Detailed Solution: Question 3


Let AB be the building and CD be the tower.
Draw BE perpendicular to CD.
 At that point CE = AB = 10m, ∠EBD = 60° and ∠ACB = ∠ CBE = 45°
AC/AB = cot45° = 1 = >AC/10 = 1 => AC = 10m.
From △ EBD, we have
DE/BE = tan 60° = √3 => DE/AC = √3
=> DE/10 = 1.732 => DE = 17.3
Height of the tower = CD = CE + DE = (10 + 17.32) = 27.3 m.

MCQ: Height & Distance - 2 - Question 4
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A man standing at a point P is watching the top of a tower, which makes an angle of elevation of 30º with the man's eye. The man walks some distance towards the tower to watch its top and the angle of the elevation becomes 60º. What is the distance between the base of the tower and the point P?

Detailed Solution: Question 4

One of AB, AD and CD must have given.

So, the data is inadequate.

MCQ: Height & Distance - 2 - Question 5
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The horizontal distance between two towers is 90 m. The angular depression of the top of the first as seen from the top of the second which is 180 m high is 450. Then the height of the first is

Detailed Solution: Question 5


=> (180 - h)/90 = Tan(45)
=> h = 90 m

MCQ: Height & Distance - 2 - Question 6
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The angle of elevation of the sun, when the length of the shadow of a tree √3 times the height of the tree, is:

Detailed Solution: Question 6

Let AB be the tree and AC be its shadow.

Let ACB = θ.
Then, AC = 3          cot θ = 3AB

 θ = 30º.

MCQ: Height & Distance - 2 - Question 7
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A man is watching form the top of the tower a boat speeding away from the tower. The boat makes the angle of depression of 60° with the man's eye when at a distance of 75 meters from the tower. After 10 seconds the angle of depression becomes 45°. What is the approximate speed of the boat, assuming that it is running in still water?

Detailed Solution: Question 7


Let AB be the tower and C and D be the positions of the boat.
Distance travelled by boat = CD
From the figure 75tan(60) = (75 + CD)tan(45)
=> 75√3 = 75 + CD
=> CD = 55 m
Speed = distance/time = 55/10
= 5.5 m/sec = 19.8 kmph

MCQ: Height & Distance - 2 - Question 8
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From a point P on a level ground, the angle of elevation of the top tower is 30º. If the tower is 100 m high, the distance of point P from the foot of the tower is:

Detailed Solution: Question 8

Let AB be the tower.

Then, APB = 30º and AB = 100 m.

 AP= (AB x √3) m
= 100√3 m
= (100 x 1.73) m
= 173 m.

MCQ: Height & Distance - 2 - Question 9
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An observer 1.4 m tall is 10√3 away from a tower. The angle of elevation from his eye to the top of the tower is 30°. The heights of the tower is

Detailed Solution: Question 9


Let AB be the observer and CD be the tower.
Then, CE = AB = 1.4 m,
BE = AC = 10v3 m.
DE/BE = Tan (30°) = 1/√3
DE = 10√3/√3 = 10
CD = CE + DE = 1.4 + 10 = 11.4 m

MCQ: Height & Distance - 2 - Question 10
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The Top of a 25 meter high tower makes an angle of elevation of 450 with the bottom of an electric pole and angle of elevation of 30 degree with the top of pole. Find the height of the electric pole.

Detailed Solution: Question 10


Let AB be the tower and CD be the electric pole. 
From the figure CA = DE
=> 25/(Tan(45)) = (25-h)/(Tan(30))
=> 25  Tan(30) = 25 - h
=> h = 25 - 25Tan(30)
= 25(1 - Tan(30)) 
= 25((√3 - 1)/√3)

MCQ: Height & Distance - 2 - Question 11
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From a point P on a level ground, the angle of elevation of the top tower is 60°. If the tower is 180 m high, the distance of point P from the foot of the tower is

Detailed Solution: Question 11


From ∠APB = 60° and AB = 180 m.
AB/AP = tan 60° =√3
AP = AB/√3 = 180/√3 = 60√3

MCQ: Height & Distance - 2 - Question 12
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The heights of two towers are 90 meters and 45 meters. The line joining their tops make an angle 450 with the horizontal then the distance between the two towers is

Detailed Solution: Question 12


Let the distance between the towers be X 
From the right angled triangle CFD 
Tan(45) =  (90 - 45)/X   
=> x = 45 meters

MCQ: Height & Distance - 2 - Question 13
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The angles of elevation of the tops of two vertical towers as seen from the middle point of the lines joining the foot of the towers are 45° & 60°.The ratio of the height of the towers is

Detailed Solution: Question 13


Tan(60) = h1/AB
=> h1 = √3AB
Tan(45) = h1/BC
=> h2 = BC
h1/ h2 = √3/1
=> h1 : h2 = √3 : 1

MCQ: Height & Distance - 2 - Question 14
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On the level ground, the angle of elevation of the top of a tower is 30°.on moving 20 meters nearer, the angle of elevation is 60°. Then the height of the tower is

Detailed Solution: Question 14


Let h be the height of tower
From figure.
20 = h (  cot30 - cot60)    
20 = h (√3 - 1/√3) 
=> 20√3 = h (3 - 1) 
=> h = 10√3.

MCQ: Height & Distance - 2 - Question 15
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The angle of elevation of a tower at a point 90 m from it is cot-1(4/5). Then the height of the tower is

Detailed Solution: Question 15


Let cot-1(4/5) = x
=> cot x = 4/5
=> tan(x) = 5/4
From the right angled triangle
Tan(x) = h/90
=> h = 5/4 * 90 =112.5 m

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About this SSC CGL Practice Test

This test contains 15 MCQs on MCQ: Height & Distance - 2 with detailed solutions for each question. It is part of the Quantitative Aptitude for SSC CGL course on EduRev, designed to align with the current SSC CGL syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

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