Aptitude questions on Height and Distance-Interview written test questions

INTRODUCTION:
The questions based on height & distance are based on the following factors:
The angle of elevation or depression.
One of the side out of height and distance.

Generally the angle of elevation or depression which are coming in different examinations are based on 300, 450, & 600.

When the angle of elevation or depression is 300 or 600:
Take the side infront of 300 as 1 part.
Take the side infront of 600 as P 3 parts.
Take the side infront of 900 as 2 parts.
By taking the ratio we can find the asking one

When the angle of elevation or depression is 450 :
Take the side infront of 450 as 1/P 2 part.
Take the side infront of 900 as 1part.
By taking the ratio we can find the asking one

QUESTION:
A person standing on the bank of the river observes that the angle subtended by a tree on the opposite bank is 60º.When he retires 100 m from the bank he finds the tree to be 30º. Find the height of the tree and the breadth of the river.

SOLUTION:
In D ABC, the side opposite to 30º which is the breadth of the river is 1 part
Let i.e. x meter
Then the side opposite to 60ºwhich is the height of tree is Ö 3 part = Ö 3x
Again in case of D ABD the side opposite to 30º which is the height of the tree as 1 part=Ö 3x And the side opposite tom 60 which is Ö 3parts=3x
3x=x+100
Therefore x i.e. breadth of the river= 50 m
And the height of the tree which is Ö 3x=50Ö 3m

QUESTION:
An observer on the top of a cliff 200 m above the sea level observes the angles of depression of the two ships on opposite sides of the cliff to be 45 and 30 respectively. Find the distance between the ships if the line joining them passes through the base of the cliff.

Solution:
In D ABD AD = BD
(as the corresponding angle 45)= 200 m
So AD = DC
Again in ADC the side opposite to 300 which is 1 part i.e. 200m
Therefore the side opposite to60 which is CD i.e. Ö 3 parts = 200Ö 3 m
The distance between the two
ships= (200 + 200Ö 3) m
= 200(1+Ö 3) m
= 200(1+1.732) m
= 5464 m
QUESTION:
A lizard is moving up a wall. From the foot of the wall at a certain distance the angle of elevation of dog’s eye with the lizard is 30.When the lizard reaches the top of the wall, the dog is 15 m away from the wall and at this point the angle of elevation of the dog’s eye with the lizard is 60.During this time if the lizard has covered 5Ö 3 m of height, the dog has covered how many meters of distance?

Solution:
Let initially the lizard is at a point E and latter it reaches at the point A
initially the dog is at a point C and latter it reaches at the point D
In D ABD, BD is opposite to angle 300 which is 1 part is 15 m
So the side opposite to 60 which is AB i.e. Ö 3 parts is 15 Ö 3m
Therefore BE = 10Ö 3 m
Then in D BCE, BE which is opposite to 30 i.e. 1 part is 10Ö 3 m
The distance BC which is opposite to 60 i.e. Ö 3 parts is 30 m.
The distance covered by the dog is CD = (30 -15) m
= 15 m


QUESTION:
Two unequal poles are erected on either side of the road of width 60 m .A man is standing at the exact middle position at which the angle of elevations are 30º and 60º respectively. Then find the
height of the shorter pole
height of the longer pole
Distance between the topmost points of the two poles

Solution:
(i) In case of D ABC the side BC is opposite to the angle
60º which is Ö 3 parts = 20 m
Therefore the height of the shorter pole i.e. AB opposite to the angle 30º
which is 1 part = 30/Ö 3 = 10Ö 3 m
(ii) In case of D CDE, CD is opposite to the angle 30º which is 1 part = 30 m
The side opposite to the angle 30º which is 1 part = 30 m
Therefore the side i.e. DE the height of the longer pole is opposite to 60º is Ö 3 parts= 30Ö 3m

(iii) In case of D ACE,Ð ACE = 90º so it is a right angled triangle.
In this case CE = opposite of angle of 900
From D CDE i.e. 2 parts = 60 m
Similarly AC is opposite side of the angle of 900 from D ABC i.e. 2 parts= 20 Ö 3 m
So the ratio between the side AC & CE = 1:Ö 3 and in this case 1 part = 20Ö 3 m
Therefore the distance between the topmost points of two poles AE which is opposite to the angle of 90º is 2 parts = 40Ö 3 m.


QUESTION:
If the shadow of a tower is 30 m when the sun’s altitude is 30º, what is the length of the shadow when the sun’s altitude is 60º?

Solution:
In D ABC, BC is opposite to the angle of 600 i.e. Ö 3 parts = 30 m
The height of the tower AB is opposite to the angle of 30º i.e.
1 part = 30/Ö 3 = 10Ö 3m
In case of D ABD,AB is opposite of 60º i.e. Ö 3 parts= 10Ö 3 m
So the length of shadow BD which is opposite to 300 is 1 part = 10m

QUESTION:
A man on the top of a vertical light house observes a boat is coming directly towards it. If it takes 10 minutes for the angle of depression to change from 30º to 60º how soon will it reach the light house?

Solution:
In D ABC let the length of AB = x meter
AB is opposite to 30º i.e. 1 part = x meters
So BC is opposite to 60º
i.e. Ö 3 parts = Ö 3 x meters
Again in D ABD AB opposite of 60º
which is Ö 3 parts= x meters
BD which one is opposite to 30º
i.e. 1 part = x/Ö 3 m
CD= (Ö 3 x-x/Ö 3)m = 2 x/Ö 3 m
The ship covers 2 x/Ö 3 m in 10 minutes


QUESTION:

A man in a boat being rowed away from a cliff 120(Ö 3+1)m high takes two minutes to change the angle of elevation of the top of the cliff from 45º to 30º.Find the speed of the boat in km/hr.


Solution:

In D ABD AB = BD =120(Ö 3+1)m
Again in D ABC AB is opposite
to 30º is 1 part = 120(Ö 3+1)
BC which is opposite to 60º is
Ö 3 parts= Ö 3 [120(Ö 3+1)]
SO that DC=Ö 3 [120(Ö 3+1)]-120(Ö 3+1)
=240m
There fore speed of boat is 240/120=2m/s
Which one is 7.2km/h


QUESTION:
From the top of a cliff150 m high, the angles of depression of two boats which are due north of the observer are 600 and 300.Find the distance between them.
Solution:
In the D ABC, the side opposite to 600 is Ö 3 parts = 150 m
The side opposite to 300 which is 1 part = 150/Ö 3 = 50Ö 3 m
Again in the D ABD the side opposite to 300 is 1 part=150 m
So the side opposite to 600 i.e. BD = Ö 3 parts which is 150Ö 3 m
DB=150Ö 3 m and BC = 50Ö 3 m
Therefore the distance between the two ships is
BD – BC=100Ö 3 m

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