This set of Aptitude Questions and Answers (MCQs) focuses on “Two Observers – Set 2”.
1. Two poles are x m apart and the height of one is double of the other. If from the mid – point between the two, an observer finds an elevation of their tops to be complementary, then what will be the height of the shorter pole?
a) x / √2 m
b) x√2 m
c) x / 2√2 m
d) x / 2√3 m
View Answer
Explanation: Let the height of shorter pole = h m, longer pole = 2h m.
Let the angle of elevation of shorter pole = y degree, then for longer pole = 90 – y degree.
➩ tan y = height / base distance
➩ tan y = h / (x / 2)
➩ tan y = 2h / x
Also,
➩ cot (90 – y) = base distance / height
➩ cot (90 – y) = (x / 2) / 2h
➩ cot (90 – y) = x / 4h
As the angles are complementary:
➩ 2h / x = x / 4h
➩ 8h2 = x2
➩ h = x / 2√2 m
2. If the angles of elevation of a pole from two points located in the same direction at distances of x and y (x > y) from its foot are 30 and 60 degree, then what will be the height of this pole?
a) √xy m
b) x√y m
c) √ (x + y)
d) √ (x – y) m
View Answer
Explanation: Let the height of the pole = h m.
➩ tan 30 = height / base distance
➩ 1 / √3 = h / x
Also,
➩ tan 60 = height / base distance
➩ √3 = h / y
On multiplying the two equations,
➩ 1 / √3 * √3 = h / x * h / y
➩ 1 = h2 / xy
➩ h = √xy m
3. From the top of a platform 40 m high the angle of depression and the angle of elevation of a building are found to be equal. What is the height of this building?
a) 76 m
b) 80 m
c) 85 m
d) 90 m
View Answer
Explanation: Let the height of the building = h m, base distance = b m, angle = a degree.
Angle of elevation
➩ tan a = height / base distance
➩ tan a = h – 40 / b
Angle of depression
➩ tan a = height / base distance
➩ tan a = 40 / b
On equating both the equations:
➩ h – 40 / b = 40 / b
➩ h = 80 m
4. It has been observed that on moving a distance of z m towards a house, the angle of elevation of its top changes from 30 to 60 degree, then what will be the height of the house?
a) z√3 / 2 m
b) z√2 / 3 m
c) z√3 m
d) 2z / √3 m
View Answer
Explanation: Let the height of the house = h m, base distance of point of 60 – degree elevation = x m.
Angle of elevation = 60 – degree
➩ tan 60 = height / base distance
➩ √3 = h / x
➩ x = h / √3 m.
Angle of elevation = 30 – degree
➩ tan 30 = height / base distance
➩ 1 / √3 = h / x + z
➩ x + z = h√3
➩ x = h√3 – z
On equating the above equations;
➩ h / √3 = h√3 – z
➩ z = h (√3 – (1 / √3))
➩ z = 2h / √3
➩ h = z√3 / 2 m
5. The tops of two poles of height 30 m and 25 m are connected by a wire. If the wire makes an angle of 30 degree with the horizontal, then what will be the length of the wire used?
a) 15 m
b) 12 m
c) 10 m
d) 12√3 m
View Answer
Explanation: Let the length of the wire = x m.
According to the question:
➩ sin 30 = height difference / length of wire
➩ 1 / 2 = 5 / x
➩ x = 10 m
6. The angle of elevation of the top of a lighthouse 100 m high, from two points on the ground on its opposite sides are 45 degree and 60 degree. What is the distance between these two points?
a) 200 / (1 + 1 / √3) m
b) 100 / (1 + 1 / √3) m
c) 200 (1 + 1 / √3) m
d) 100 (1 + 1 / √3) m
View Answer
Explanation: Given,
Height = 100m, point A angle = 45, point B angle = 60
Let the distance from point A to lighthouse = x m, from point B to lighthouse = y m.
Point A
➩ tan 45 = height / base distance
➩ 1 = 100 / x
➩ x = 100 m
Fish B
➩ tan 60 = height / base distance
➩ √3 = 100 / y
➩ y = 100 / √3 m
Total distance = 100 + 100 / √3
= 100 (1 + 1 / √3) m
7. Two fishes M and N are swimming away from a glacier in the same direction. If the height of the glacier is 800 m and the angle of elevation made by the 2 fishes are 45 and 30 degree respectively, then what is the distance between the two fishes?
a) 800 (√3 – 1) m
b) 800 / 2√3 m
c) 800 / (√3 – 1) m
d) 1600 / 2√3 m
View Answer
Explanation: Given,
Angle of M fish = 45 – degree, angle of N fish = 30 – degree, height = 800m
Let the distance between fish M and glacier be x m, fish N and glacier be y m.
Fish M
➩ tan 45 = height / base distance
➩ 1 = 800 / x
➩ x = 800 m
Fish N
➩ tan 30 = height / base distance
➩ 1 / √3 = 800 / y
➩ y = 800√3 m
Distance between the two ships = y – x
= 800√3 – 800 m
= 800 (√3 – 1) m
8. A building is surmounted by a vertical pole of height 20 m. From a point on the ground, the angle of elevation of the bottom and top of the pole are 30 and 60 degree respectively. What is the height of the building?
a) 10√3 m
b) 15 / √3 m
c) 10 m
d) 10 / √3 m
View Answer
Explanation: Let the base distance be = b m, height of the building = h m.
Angle = 30 degree
➩ tan 30 = height / base distance
➩ 1 / √3 = h / b
➩ b = h√3 m
Angle = 60 degree
➩ tan 60 = height / base distance
➩ √3 = h + 20 / b
➩ b = h + 20 / √3 m
On equating both the equations:
➩ h√3 = h + 20 / √3
➩ 3h = h + 20
➩ 2h = 20
➩ h = 10 m
9. If the length of a shadow cast by a pole is √3 times the length of the object, then what is the length of the object?
a) 53 degree
b) 60 degree
c) 30 degree
d) 37 degree
View Answer
Explanation: Let the angle be = x degree, length of object = z m, length of shadow = z√3 m.
➩ tan x = height / base
➩ tan x = z / z√3
➩ tan x = 1 / √3
➩ x = 30 degree
10. From a point G on the ground, the angle of elevation of the top of the pole is 45 degree. It the pole is 450 m high, then what will be the distance between the foot of the pole and point G?
a) 450 m
b) 425 m
c) 400 m
d) 475 m
View Answer
Explanation: Given,
Angle = 45 – degree, height = 450 m
Let the base distance be = b m.
➩ tan 45 = height / base distance
➩ 1 = 450 / b
➩ b = 450 m
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