Height and Distance- Aptitude Questions and Answers

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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 ) )

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