# Height and Distance - aptitude test

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## Height and Distance - Posted on August 02, 2013 at 01:10 PM by Rahul Choudhary

1) The angle of elevation of the sun is 60°. Find the length of the shadow of a man who is 180 cm tall.

a) 127.27 cm
b) 103.92 cm
c) 311.77 cm
d) None of these

SOLUTION : Let AB be the man and BC be his shadow
AB = 180
tan60° = AB/BC
v3 = 180/BC
BC = 103.92 cm

2) What is the distance between a tower and an observer if the angle of elevation from the observer’s eye to the top of the tower (height = 50 m) is 30°? The observer is 1.5 m tall.

a)84
b)28
c)30
d)90

SOLUTION : Let AB be the tower and CD be the observer
tan30° = AE/CE
1/v3 = (AB – CD) / x = (50 – 1.5)/ x = 48.5/x
x = 84 m

3) Two boats are spotted on the two sides of a light house. If the angle of depression made by both the boats from top of the lighthouse is 30° and 45° and the height of the light house is 125 m then find the distance between the two boats.

a)1
b)188.56 m
c)250 m
d)197.17 m

SOLUTION : Let AB be the light house and the two boats be at C and D
AB = 125 m
tan30° = BC/AB = x/125 = 1/v3
x = 72.17 m
tan45° = BD/AB = y/125 = 1
y = 125 m
Therefore, the distance between the two boats is = x + y
= 72.17 + 125
= 197.17 m

4) A man standing on the terrace of a building watches a car speeding towards him. If at that particular instant the car is 200 m away from the building makes an angle of depression of 60° with the man’s eye and after 8 seconds the angle of depression is 30°, what is the speed of the car?

a)15 m/s
b)16.67 m/s
c)25 m/s
d)Cannot be determined

SOLUTION : Let AB be the building and the man is standing at A.
When the car is 200m away from the building, the angle of depression is 60°

tan60 = BD/AB = v3

200/AB = v3

AB = 115.47 m

tan30° =BC/AB = x/115.47 = 1/v3

x = 66.67

Now, the car travels distance CD in 8 seconds

CD = BD – BC = 200 – 66.67 = 133.33 m

Speed = 133.33/8 = 16.67 m/s

5) A ship is approaching an observation tower. If the time taken by the ship to change the angle of elevation from 30° to 45° is 10 minutes, then find the time the ship will take to cover the remaining distance and reach the observation tower assuming the ship to be travelling at a uniform speed.

a) 15 minutes 20 seconds
b) 13 minutes 40 seconds
c) 16 minutes 40 seconds
d) Cannot be determined

ANSWER : 13 minutes 40 seconds

SOLUTION : Let AB be the observation tower and h be its height.

Also, let the ship be at C when the angle of elevation is 30° and at D when the angle of elevation is 45°.

The time taken by the ship to travel from C to D is 10 minutes and we need to find out the time the ship will take to reach B from D.

tan30 = AB/CB = h/CB = 1/v3

CB = v3*h

tan45 = AB/DB = h/DB = 1

DB = h

CD = CB – DB = (v3h – h) = h(v3 – 1)

Now, as h(v3 – 1) distance is covered in 10 minutes, a distance of h is covered in = 13.66 minutes = 13 minutes 40 seconds

6) On the two sides of a road are two tall buildings exactly opposite to each other. The height of the taller building is 60 m. If the angle of elevation from the top of the smaller building to the top of the taller one is 30° and the angle of depression from top of the taller building to the foot of the smaller one is 30°, then find the height of the smaller building.

a) 45
b) 30
c) 40
d) 20

SOLUTION : Let AB be the taller building of height 60m and CD be the smaller one of height h m.

tan30= DB/AB = DB/60 = 1/v3

DB = 34.64 m

tan30= AE/CE = AE/DB = 1/v3 = AE/34.64

AE = 60 – h = 20

h = 40 m

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##### A tree of height 100m has a shadow of length 100 m at a particular instant of the day. Find the angle of elevation of the sun at this point of time.

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