This set of Aptitude Questions and Answers (MCQs) focuses on “Two Observers”.

1. Two fishes are swimming in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse from the two fishes is 30 degree and 45 degree. If the lighthouse is 200 m high, what is the distance between the two fishes?

a) 200√3 m

b) 400√3 m

c) 200 (1 + √3) m

d) 400 (1 + √3) m

View Answer

Explanation: Given,

Height = 200m, fish A angle = 30, fish B angle = 45

Let the distance from fish A to lighthouse = x m, from fish B to lighthouse = y m.

Fish A

➩ tan 30 = height / base distance

➩ 1 / √3 = 200 / x

➩ x = 200√3 m

Fish B

➩ tan 45 = height / base distance

➩ 1 = 200 / y

➩ y = 200 m

Total distance = 200√3 + 200

= 200 (1 + √3) m

2. Two ships X and Y are sailing away from a glacier in the same direction. If the height of the glacier is 1000 m and the angle of elevation made by the 2 ships are 45 and 30 degree respectively, then what is the distance between the two ships?

a) 500(√3 + 1) m

b) 1000(√3 – 1) m

c) 1000(√3 + 1) m

d) 500(√3 – 1) m

View Answer

Explanation: Given,

Angle of X ship = 45 – degree, angle of Y ship = 30 – degree, height = 1000m

Let the distance between ship X and glacier be x m, ship Y and glacier be y m.

Ship X

➩ tan 45 = height / base distance

➩ 1 = 1000 / x

➩ x = 1000 m

Ship Y

➩ tan 30 = height / base distance

➩ 1 / √3 = 1000 / y

➩ y = 1000√3

Distance between the two ships = y – x

= 1000√3 – 1000 m

= 1000 (√3 – 1) m

3. Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse from the two ships is 30 degree and 60 degree. What is the relationship between the distances of the 2 ships from the light house?

a) x = 3y

b) x = y√3

c) x / √3 = 3y

d) x / √3 = y / 2

View Answer

Explanation: Given,

Angle from ship 1 = 30 – degree, ship 2 = 60 – degree

Let the base distance of ship 1 = x m, ship 2 = y m, height of lighthouse = h m.

Ship 1

➩ tan 30 = height / base distance

➩ 1 / √3 = h / x

➩ x / √3 = h

Ship 2

➩ tan 60 = height / base distance

➩ √3 = h / y

➩ h = y√3

On equating both the equations;

➩ x / √3 = y√3

➩ x = 3y

4. When the sun’s altitude changes from 30to 60 degree, the length of the shadow of a pole decreases by 100 m. What is the original length of the shadow?

a) 175 m

b) 200 m

c) 150 m

d) 125 m

View Answer

Explanation: Given,

Original angle = 30 – degree, new angle = 60 degree

Let the original length of the shadow be = x m, new length = x – 100 m, height of the pole = h m.

Original angle

➩ tan 30 = height / base length

➩ 1 / √3 = h / x

➩ x = h√3 m

New angle

➩ tan 60 = height / base length

➩ √3 = h / x – 100

➩ √3(x – 100) = h

Substituting this value of h in first equation:

➩ x = h√3

➩ x = (√3(x – 100)) * √3

➩ x = 3x – 300

➩ 300 = 2x

➩ x = 150 m

5. On the same side of a building, two objects are located. The angle of depression from the top of the building is 45 and 60 degree. If the height of the building is 250 m, what is the distance between the two objects?

a) 500(1 – 1 / √3) m

b) 250(1 – √3) m

c) 500(1 – √3) m

d) 250(1 – 1 / √3) m

View Answer

Explanation: Given,

Angle of depression for object 1 = 45 – degree, object 2 = 60 – degree, height of the building = 250 m.

Let the base distance of object 1 = x m, object 2 = y m.

Object 1

➩ tan 45 = height / base distance

➩ 1 = 250 / x

➩ x = 250 m

Object 2

➩ tan 60 = height / base distance

➩ √3 = 250 / y

➩ y = 250 / √3

The distance between the two objects = x – y

= 250 – 250 / √3

= 250 (1 – 1 / √3) m

6. The angle of elevation of a cloud from a point 100 m above a lake is 30 degree and the angle of depression of its reflection in the lake is 60 degree. What will be the height of the cloud?

a) 150 m

b) 180 m

c) 100 m

d) 120 m

View Answer

Explanation: Let the height of the cloud be h m and base distance be b m.

➩ tan 30 = height / base distance

➩ 1 / √3 = h / b

➩ b = h√3

According to the question:

➩ tan 60 = height / base distance

➩ √3 = (h + 100 + 100) / b

Substituting the value of b in terms of h from the above equation

➩ √3 = (h + 200) / h√3

➩ 3h = h + 200

➩ 2h = 200

➩ h = 100 m

7. The top of a 20 m high building makes an angle of elevation of 60 with the bottom of an electric pole and angle of elevation of 30 with the top of the pole. What is the height of the electric pole?

a) 40 / √3 m

b) 20√3 m

c) 40 / 3 m

d) 20 / 3 m

View Answer

Explanation: Let the height of the pole be h m and base distance be b m.

➩ tan 60 = height / base distance

➩ √3 = 20 / b

➩ b = 20 / √3 m

Also,

➩ tan 30 = height / base

➩ 1 / √3 = 20 – h / b

➩ b = (20 – h) √3

On equating the two equations:

➩ 20 / √3 = (20 – h) √3

➩ 20 = 3(20 – h)

➩ 20 = 60 – 3h

➩ 3h = 40

➩ h = 40 / 3 m

8. From the top of a pole, the angles of depression of two objects P and Q situated on the same side are 30 degree and 45 degree separated by a distance of 100 m. What will be the height of this pole?

a) 100 / (√3 – 1) m

b) 200√3 m

c) 200 (√3 + 1) m

d) 100 / √3 m

View Answer

Explanation: Let the height of the pole = h m, base distance of the point having 45 degree = x m.

➩ tan 45 = height / base distance

➩ 1 = h / x

➩ h = x

Also,

➩ tan 30 = height / base distance

➩ 1 / √3 = h / (x + 100)

➩ x + 100 = h√3

On substituting x from the above equation,

➩ h + 100 = h√3

➩ 100 = h (√3 – 1)

➩ h = 100 / (√3 – 1) m

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