Height and Distance- Aptitude Questions and Answers

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An aeroplane flying at a height of 300 m above the ground passes vertically above another plane at an instant when the angles of elevation of the two planes from the same point on the ground are 60º and 45º respectively. What is the height of the lower plane from the ground?

A)
100√3 m

B)
100/√3 m

C)
50/√3 m

D)
150(√3 +1) m


Correct Answer :

100√3 m


Explanation :

Let the height of the lower plane from the ground be h m and PA = x

Now, in BAP,
tan45° = AB/AP
1 = h/x
h = x ....eq(i)

Now In APC
tan60° = AC/AP = 300/x
x = 300/√3 ....eq(ii)

From Eqs. (i) and (ii), we get
h = x = 300/√3
= 300*√3 /√3*√3
= 300*√3 /3
= 100√3 m

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A boy is flying a kite. The string of the kite makes an angle of 60o with the ground. If the height of the kite is 32 m, find the length (in meters) of the string.

A)
64

B)
32/√3

C)
32√3

D)
64/√3


Correct Answer :

64/√3


Explanation :

Sin q = Perpendicular/Hypotenuse
Sin 60o = 32/ L
L=32/ √3/2

Hence, the length (in meters) of the string L= 64/√3

A pole of height h = 60 ft has a shadow of length l= 60 ft at a particular instant of time. Find the angle of elevation (in degrees) of the sun at this point of time.

A)
30

B)
45

C)
60

D)
40


Correct Answer :

45


Explanation :

let, hight of the pole = BC and length of shadow is = AB
tan A = Perpendicular/Base = BC / AB = h / l = 60 / 60 = 1
tan A = 1 for A =45o.
Hence the angle of elevation of the sun at this point of time is 45o

A person standing on bank of the river observes that the angle subtended by a tree on the opposite bank is 60° and when he retires 40m from the bank, he finds the angle of elevation as 30°. Then the breadth of the river is

A)
40m

B)
20m

C)
22m

D)
35m


Correct Answer :

20m


Explanation :


Let the height of the tree is BC and breadth of the river is AB.
there A was the initial position and D is the final potion of the man

tan 60 =h/x
√3 =h/x
h=x√3 -----------eq(1)

tan 30 = h/ 40+x
1/√3 = h/ 40+x
h=(40+x)/√3

using eq(1)
x√3=(40+x)/√3
3x = 40+x
2x = 40
x = 20

Two men on one sides of a building of height 25 m notice the angle of elevation of the top of the building to be 45o and 60o respectively. Find the distance (in meters) between the two men.

A)
32(1+ (1/√3))

B)
15/√3

C)
25(1-(1/√3))

D)
25√3


Correct Answer :

25(1-(1/√3))


Explanation :


Let In the figure, A and B represent the two men and CD the tall building.
tan A = tan 45= h / AC
tan B = tan 60 = h / BC.
Now the distance between the men is AB
= x = AC − BC = (h / tan 45) − (h / tan 60)
= h((1/ tan 45) − (1/ tan 60 ) )
= 25(1 − (1/ tan 60 ) )
= 25(1 − (1/√3 ) )

From the top of a building 90 m high, the angles of depression of the top and the bottom of a tree are 30º and 45º respectively. What is the height of the tree?

A)
30√3 m

B)
90-30√3 m

C)
90+30√3 m

D)
60+30√3 m


Correct Answer :

90-30√3 m


Explanation :

Let AB and CD be the building and tree respectively.
Now, in ABD
tan45° = AB/BD
1 = 90/BD
BD = 90 m ...eq(i)

Again, in AEC
tan30° = AE/EC
1/√3 = AE/90
AE = 90/√3 = 30√3 m

Hence, Height of tree = CD = BE = AB-AE = 90-30√3 m

The difference in the length of the shadow of a 10 m high pole when the angles of elevation of the sun are 30° and 60° is

A)
20√3 / 3 m

B)
10√3 m

C)
10√3 / 3 m

D)
10 m


Correct Answer :

20√3 / 3 m


Explanation :


Let AB be the pole and BD, BC is its shadows when the angle of elevation of the sun is 30°, 60° respectively.

in ACB
tan 60° = √3 = AB/BC = 10/BC
BC = 10/√3 = 10√3 / 3

in ADB
tan 30° = 1/√3 = AB/BD = 10/ BC + CD
BC + CD = 10√3
CD = 10√3 - 10√3/3 =(30√3 - 10√3) / 3 = 20√3 / 3

The difference between the length of the shadow, CD = 20√3 / 3

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