All Courses
Get the App Signup / Login
Sign in Open App
SSC CGL Exam  >  SSC CGL Tests  >  SSC CGL Tier 2 - Study Material, Online Tests, Previous Year  >  MCQ: Height & Distance - 2 - SSC CGL MCQ

MCQ: Height & Distance - 2 - SSC CGL MCQ


Test Description

15 Questions MCQ Test SSC CGL Tier 2 - Study Material, Online Tests, Previous Year - MCQ: Height & Distance - 2

MCQ: Height & Distance - 2 for SSC CGL 2024 is part of SSC CGL Tier 2 - Study Material, Online Tests, Previous Year preparation. The MCQ: Height & Distance - 2 questions and answers have been prepared according to the SSC CGL exam syllabus.The MCQ: Height & Distance - 2 MCQs are made for SSC CGL 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for MCQ: Height & Distance - 2 below.
Solutions of MCQ: Height & Distance - 2 questions in English are available as part of our SSC CGL Tier 2 - Study Material, Online Tests, Previous Year for SSC CGL & MCQ: Height & Distance - 2 solutions in Hindi for SSC CGL Tier 2 - Study Material, Online Tests, Previous Year course. Download more important topics, notes, lectures and mock test series for SSC CGL Exam by signing up for free. Attempt MCQ: Height & Distance - 2 | 15 questions in 15 minutes | Mock test for SSC CGL preparation | Free important questions MCQ to study SSC CGL Tier 2 - Study Material, Online Tests, Previous Year for SSC CGL Exam | Download free PDF with solutions
MCQ: Height & Distance - 2 - Question 1
Save

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 for MCQ: Height & Distance - 2 - 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
Save

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 for MCQ: Height & Distance - 2 - 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.

1 Crore+ students have signed up on EduRev. Have you? Download the App
MCQ: Height & Distance - 2 - Question 3
Save

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 for MCQ: Height & Distance - 2 - 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
Save

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 for MCQ: Height & Distance - 2 - Question 4

One of AB, AD and CD must have given.

So, the data is inadequate.

MCQ: Height & Distance - 2 - Question 5
Save

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 for MCQ: Height & Distance - 2 - Question 5


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

MCQ: Height & Distance - 2 - Question 6
Save

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 for MCQ: Height & Distance - 2 - 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
Save

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 for MCQ: Height & Distance - 2 - 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
Save

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 for MCQ: Height & Distance - 2 - 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
Save

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 60°. The heights of the tower is
Detailed Solution for MCQ: Height & Distance - 2 - 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
Save

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 for MCQ: Height & Distance - 2 - 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
Save

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 for MCQ: Height & Distance - 2 - 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
Save

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 for MCQ: Height & Distance - 2 - 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
Save

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 for MCQ: Height & Distance - 2 - 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
Save

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 for MCQ: Height & Distance - 2 - 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
Save

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 for MCQ: Height & Distance - 2 - 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

1365 videos|1312 docs|1010 tests
Join Course
Information about MCQ: Height & Distance - 2 Page
In this test you can find the Exam questions for MCQ: Height & Distance - 2 solved & explained in the simplest way possible. Besides giving Questions and answers for MCQ: Height & Distance - 2, EduRev gives you an ample number of Online tests for practice

Top Courses for SSC CGL

Download as PDF

Top Courses for SSC CGL

Important Questions for MCQ: Height & Distance - 2

Find all the important questions for MCQ: Height & Distance - 2 at EduRev.Get fully prepared for MCQ: Height & Distance - 2 with EduRev's comprehensive question bank and test resources. Our platform offers a diverse range of question papers covering various topics within the MCQ: Height & Distance - 2 syllabus. Whether you need to review specific subjects or assess your overall readiness, EduRev has you covered. The questions are designed to challenge you and help you gain confidence in tackling the actual exam. Maximize your chances of success by utilizing EduRev's extensive collection of MCQ: Height & Distance - 2 resources.

MCQ: Height & Distance - 2

Prepare for the MCQ: Height & Distance - 2 within the SSC CGL exam with comprehensive MCQs and answers at EduRev. Our platform offers a wide range of practice papers, question papers, and mock tests to familiarize you with the exam pattern and syllabus. Access the best books, study materials, and notes curated by toppers to enhance your preparation. Stay updated with the exam date and receive expert preparation tips and paper analysis. Visit EduRev's official website today and access a wealth of videos and coaching resources to excel in your exam.

Online Tests for MCQ: Height & Distance - 2 SSC CGL Tier 2 - Study Material, Online Tests, Previous Year

Practice with a wide array of question papers that follow the exam pattern and syllabus. Our platform offers a user-friendly interface, allowing you to track your progress and identify areas for improvement. Access detailed solutions and explanations for each test to enhance your understanding of concepts. With EduRev's Online Tests, you can build confidence, boost your performance, and ace MCQ: Height & Distance - 2 with ease. Join thousands of successful students who have benefited from our trusted online resources.
© EduRev
Education Revolution
Signup to see your scores go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup