Short Tricks on Height & Distance
Table of Contents
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.
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.
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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.
Numerically angle of elevation is equal to the angle of depression.
Both the angles are measured with the horizontal.
Previous year Questions based on Height & Distance asked in SSC CGL Exam and SSC CGL Tier II Exam.
- 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:
Sin 30° = h/120
1/2 = h/120
h = 60m
- 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:
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)
- 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:
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
- 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:
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)
- 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:
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
Important short trick are :
Note: only when the sum of angle i.e
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 :
(c)500
Ans. (d)
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
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 45^{0} 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 &
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 30^{0}. After some time, the angle of depression becomes 60^{0}. The distance (in m) travelled by the car during this time is –
Ans. (c)
Solution:
AB = AC – BC
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 60^{0}. When he moves 50m away from the bank, the angle of elevation becomes 30^{0}. The height of the tree and width of river respectively are :
(c)
(d) None of these
Answer: a)
Solution:
Ratio value original value
height of the tree= h (ratio value = )=
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 30^{0} and 45^{0} respectively. The height of second pillar (in metre) is:
Answer: (c)
Solution:
Let AB and CD are pillars.
Let DE = h
Required height
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 30^{0} and 60^{0}. The height of each pole is:
Ans. (d)
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 30^{0} and 60^{0} respectively. The height of the other post is :
(a)36
(b)72
(c)76
(d)80
Ans (b)
The height of greater Lower i.e. AB = 108 = H
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 60^{0} and 45^{0} respectively. The vertical distance between the aeroplanes at that instant is:
(d)4500 m
Ans (c)
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 30^{0} and after 2 minutes, he observes the same bird in the south at an angle of elevation of 60^{0}. 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)
In ABO
According to ratio method
From triangle DCO
DO cot AO = 150 + 50 = 200 m
(5)A tree is broken by the wind. If the top of the tree struck the round at an angle of 30^{0} and at a distance of 30 m from the root, then the height of the tree is :
Ans. (b)
√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 :
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
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
From equations (i) and (ii),
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