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Short Tricks on Height & Distance

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Short Tricks on Height & Distance

Today we will be covering a very important topic from the Advance Maths part of the Quantitative Aptitude section that is – Important Notes & Short Tricks on Height & Distance.

Important Notes & Short Tricks on Height & Distance

Short Tricks on Height & Distance

Angle of Elevation: Let AB be a tower/pillar/shell/minar/pole etc.) standing at any point C on the level ground is viewing at A.

Important Notes & Short Tricks on Height & Distance

The angle ,which the line AC makes with the horizontal line BC is called angle of elevation .so angle ACB is angle of elevation.

Angle of Depression: If observer is at Q and is viewing an object R on the ground ,then angle between PQ and QR is the angle of depression .so angle PQR is angle of depression.

Important Notes & Short Tricks on Height & Distance

Numerically angle of elevation is equal to the angle of depression.

Both the angles are measured with the horizontal.

Important Notes & Short Tricks on Height & Distance

Previous year Questions based on Height & Distance asked in SSC CGL Exam and SSC CGL Tier II Exam.

  1. The thread of a kite is 120 m long and it is making 30° angular elevation with the ground .What is the height of the kite?

Solution:

Important Notes & Short Tricks on Height & Distance

Sin 30° = h/120

1/2 = h/120

h = 60m

 

  1. A tree bent by the wind .The top of the tree meets the ground at an angle of 60°.If the distance between the top of the foot be 8 m then what was the height of the tree?

Solution:

Important Notes & Short Tricks on Height & Distance

tan 60° = x/8

√3 = x/8

x = 8 √3

y cos 60° = 8/y

1/2 = 8/y

y = 16

therefore height of the tree = x+y

= 8√3+16

= 8(√3+2)

  1. The angle of elevation of the top of a tower from a point on the ground is 30° . On walking 100m towards the tower the angle of elevation changes to 60° . Find the height of the tower.

Solution:

Important Notes & Short Tricks on Height & Distance

In right triangle ABD,

tan 60° = h/x

√3 x = h

x = h/√3

Again , in right triangle ABC ,

tan 30 = h/x+100

1/√3 = h/x+100

√3 h = x+100

√3 h = h/√3 + 100

√3 h – h/√3 =100

3 h – h/√3    =100

2 h = 100√3

h = 50√3

By short trick:

d = h (cot Ɵ1 – cot Ɵ2)

h = 100/(√3-1/√3) = 100*√3/2 = 50√3

Ɵ1 = small angle

Ɵ2 = large angle

d = distance between two places

h = height

  1. From the top of a temple near a river the angles of depression of both the banks of river are 45° & 30°. If the height of the temple is 100 m then find out the width of the river.

Solution:

Important Notes & Short Tricks on Height & Distance

tan 45° = AB/BD

1 = 100/BD

BD = 100

tan 30 ° = AB/BC

1/√3 = 100/BC

BC = 100 √3

Width of the river , CD = BC – BD = 100 (√3-1)

When height of tower is 1 m then width of river is √3-1

Since height of tower is 100 m

Therefore ,

Width of river is 100(√3-1)m

By short trick:

Same formula can be used in this question too i.e.

d= h (cot Ɵ1 – cot Ɵ2)

  1. The angle of elevation of the top of a tower from a point is 30 °. On walking 40 m towards the tower the angle changes to 45°.Find the height of the tower?

Solution:

Important Notes & Short Tricks on Height & Distance

 

tan 45° = AB/BD

1 = AB/1

Therefore AB = 1

tan 30° = AB/BC =>1/√3 = 1/BC

therefore BC= √3

Now CD =√3-1 m and height of tower is 1 m

1 m = 1/√3-1

Therefore 40 m  = 1/√3-1.40 = 40/√3-1

= 20 (√3+1)m

By trick:

40 = h(√3-1)

H  = 40/(√3-1) = 20 (√3+1)m

Height & Distance Part -2

In this part we will try to solve question with the help of ratio. This trick will save a lot of time and useful during the exam period.  This article also contains some  useful formula .

Here are some ratio figure which you have to remember

111

 

 

Important short trick are :

image049

Noteimage050 only when the sum of angle i.e  image051

 

ssc height and distance 1

ssc height and distance

Some Important question are as follows:

Example 1:The angle of elevation of the top of a tower at a distance of 500 m from its foot is 30°. The height of tower is :

(a)image014

(b)image016

(c)500

(d)image015

Ans. (d)

image017

Short trick:

Solve it with ratio , as the angle of elevation is 30° then ratio between P: B: H  is 1:√3:2 so √3= 500 then 1= 500/√3 and height is equal to image015

Example 2: The banks of a river are parallel. A swimmer starts from a point on one of the banks and swims in a straight line inclined to the bank at 450 and reaches the opposite bank at a point 20 m from the point opposite to the starting point. The breadth of the river is :

(a)  20 m

(b)  28.28 m

(c)   14.14 m

(d)  40 m

Ans. (c) 14.14 m

Solution:

Let A be the starting point and B, the end point of the swimmer. Then AB = 20m &  image018

image019

image021

image020

Short Method;

AS the angle of elevation is 45° then the ratio of P: B : H i.e. 1:1:√2

here √2 =20 then 1 =20/√2

Question 3: A man from the top a 50m high tower, sees a car moving towards the tower at an angle of depression of 300. After some time, the angle of depression becomes 600. The distance (in m) travelled by the car during this time is –

(a)image022

(b)image023

(c)image024

(d)image025

Ans. (c)

Solution: 

image026

AB = AC – BC

image029

image030

Example 4:A person standing on the bank of a river observes that the angle of elevation of the top of a tree on the opposite side of the bank is 600. When he moves 50m away from the bank, the angle of elevation becomes 300. The height of the tree and width of river respectively are :

(a) image033

(b)  image032

(c) image031

(d)  None of these

Answer: a)

Solution: 

image034

 

image035

 

Ratio value original value

image036image037

 

height of the tree= h (ratio value = image039)=image040

and width of the river = x (ratio value = 1) = 25 m

Example 5: From the top of a pillar of height 80 m the angle of elevation and depression of the top and bottom of another pillar are 300 and 450 respectively. The height of second pillar (in metre) is:

(a)   image041 m

(b)  image042

(c)    image043

(d) image044

Answer: (c)

Solution: 

Let AB and CD are pillars.

Let DE = h

image045

 

In image046

 

image047

Required height

image048

Height & Distance Part -3

In this part we will try to solve question with the help of ratio. This part include some important question for the SSC Exam.

(1)Two poles of equal height are standing opposite to each other on either side of a road, which is 28m wide. From a point between them on the road, the angles of elevation of the tops are 300 and 600. The height of each pole is:

(a)image001

(b)image002

(c)image003

(d)image009

 

Ans.   (d)

image005

Let AB and CD be the pole and AC be the road.

Let AE = x, then EC = 28-x and AB = CD = h. Then   let AB = CD=√3

then, EC =1 and AE = 3

AC (ratio value) = 3 + 1 = 4

4 = 28 then 1 =7

and √3=7√3 so height of tower is 7√3.

(2)There are two vertical posts, one on each side of a road, just opposite to each other. One post is 108 metre high. From the top of this post, the angles of depression of the top and foot of the other post are 300 and 600 respectively. The height of the other post is :

(a)36

(b)72

(c)76

(d)80

Ans (b)

image010

The height of greater Lower i.e.  AB = 108 = H

image011

so height of tower is 72

(3)An aeroplane when flying at 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 600 and 450 respectively. The vertical distance between the aeroplanes at that instant is:

(a)image012

(b)image013

(c)image014

(d)4500 m

Ans (c)

image015

image016

In this question we have two triangle ABC and triangle DBC. In triangle ABC we apply the ratio according to 60° and in triangle DBC we apply ratio according to the 45°. That why we take AB=√3 and DB =1.

(4)A boy standing in the middle of a field, observes a flying bird in the north at an angle of elevation of 300 and after 2 minutes, he observes the same bird in the south at an angle of elevation of 600. If the bird flies all along in a straight line at a height of then its speed in km/h is:

(a)  4.5

(b)  3

(c)   9

(d)  6

Ans.(d)

image018

In  ABO

According to ratio method

image020

From triangle DCO

image022

DO cot AO = 150 + 50 = 200 m

Speed = image023

 

image024

(5)A tree is broken by the wind. If the top of the tree struck the round at an angle of 300 and at a distance of 30 m from the root, then the height of the tree is :

(a) image025

(b) image026

(c) image027

(d) image028

Ans. (b) 

image029

√3=30

1= 10√3  & 2 =20√3

so total height is 1+2 =10√3+20√3= 30√3

(6)The angle of elevation of a cloud from height h above the level of water in a lake is a and the angle of the depression of its image in the lake is b. Then, the height of the cloud above the surface of the lake is :

(a)   image033

(b)image034

(c)  image035

(d) image036

Ans. (d)

Let P be the cloud at height H above the level of the water in the lake Q its image in the water

image037

image038,

B is at a point at a height AB = h, above the water, Angle of elevation of P and depression of Q from B are respectively

In  traingle PBM

image040

From equations (i) and (ii),

image041

 

 

 

 

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