^{0}. The man walks some distance to wards the tower and then this angle of elevation of the top of the tower is 60^{0}. If the height of tower is 30m, then the distance he moves is

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The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it is 60°. Find the height of the tower.

A.

10√3 m

B.

20√3 m

C.

30√3 m

D.

40√3 m

Answer with explanation

Answer: Option BExplanation

AB is the tower and BC is the length of the shadow when the Sun’s altitude is 60°,

i.e., the angle of elevation of the top of the tower from the tip of the shadow is 60° and DB is the length of the shadow, when the angle of elevation is 30°.

Now, let AB be h m and BC be x m.

According to the question, DB is 40 m longer than BC.

So, DB = (40 + x) m

Now, we have two right triangles ABC and ABD.

In Δ ABC,

tan 60° = AB / BC

or, √3 = h / x ….. (1)

In Δ ABD,

tan 30° = AB / BD

i.e., 1 / √3 = h / (40 + x ) ….. (2)

From (1), we have

h = x√3

Putting this value in (2), we get

(x√3)√3 = x + 40,

i.e., 3x = x + 40

i.e., x = 20

Workspace

Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30° and 45° respectively. If the lighthouse is 100

A.

173 m

B.

200 m

C.

273 m

D.

300 m

Answer with explanation

Answer: Option CExplanation

Let AB be the lighthouse and C and D be the positions of the ships

.Then, AB = 100 m, ACB = 30° and ADB = 45°

.AB= tan 30° =1 AC = AB x 3 = 1003 m.AC3AB= tan 45° = 1 AD = AB = 100 m.AD CD = (AC + AD)= (1003 + 100) m= 100(3 + 1)= (100 x 2.73) m= 273 m.

Workspace

From a point C on a level ground, the angle of elevation of the top of a tower is 30 degree. If the tower is 100 meter high, find the distance from point C to the foot of the tower.

A.

170 meter

B.

172 meter

C.

173 meter

D.

167 meter

Answer with explanation

Answer: Option CExplanation

Let AB be the tower.

then∠ACB=30∘AB=100meterABAC=tan30∘=>100AC=13=>AC=3∗100=>AC=1.73∗100=>AC=173m

Please always remember the value of square root 3 is 1.73, and value of square root 2 is 1.41.

This will be very helpful while solving height and distance questions and saving your time.

Workspace

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

Answer with explanation

Answer: Option BExplanation

Let AB be the man and BC be his shadow

AB = 180

tan60° = AB/BC

v3 = 180/BC

BC = 103.92 cm

Workspace

40 feet rope is cut into 2. One piece is 18 feet longer than the other. What is the length of the shorter piece?

A.

11

B.

12

C.

18

D.

22

Answer with explanation

Answer: Option AExplanation

We can also test the answer choices

Start with the middle answer choice C

C) 18

If the shorter piece is 18 feet long, then the longer piece is 36 feet long

Sum of lengths = 18 + 36 = 54

NO GOOD. We want a sum of 40 feet (as per the question)

So, we can ELIMINATE C.

Also, since we need the two pieces to be SHORTER, we can also ELIMINATE D and E

Now try answer choice B

B) 11

If the shorter piece is 11 feet long, then the longer piece is 29 feet long

Sum of lengths = 11 + 29 = 40

PERFECT!!!

Answer: B

Workspace

^{0}. The man walks some distance to wards the tower and then this angle of elevation of the top of the tower is 60^{0}. If the height of tower is 30m, then the distance he moves is

A.

22 m

B.

22√3 m

C.

20 m

D.

20√3m

Answer with explanation

Answer: Option BExplanation

Workspace

A.

5√3 meter

B.

5(3-√3) m

C.

24√2 meter

D.

96 meter

Answer with explanation

Answer: Option BExplanation

Since, by the below diagram,

We can write,

Also, Again by the diagram,

Hence, the distance between A and B is,

Hence, **The distance between the points is 5(3-√3) unit**

Workspace

A man 6ft tall cats a shadow 4ft long. At the same time when a flag pole casts a shadow 50 ft long. The height of the flag pole is

A.

15 ft

B.

75 ft

C.

60 ft

D.

70 ft

Answer with explanation

Answer: Option BExplanation

**Let x represent how tall the flagpole is
Question Sets Up the Proportion
**

** x/10=6/4
Cross Multiplying
4x = 60
x = 15ft tall**

** **

Workspace

A man is standing on the deck of a ship, which is 10m above water level. He observes the angle of elevation of the top of a light house as 600 and the angle of depression of the base of lighthouse as 300. Find the height of the light house.

A.

30m

B.

40m

C.

45m

D.

38m

Answer with explanation

Answer: Option BExplanation

from A is 60 °

Then ∠EAD= 60° &

∠CAE= ∠BCA= 30°. (Alternate ANGLES)

Let AD = BC = x m & DE= h m

In ∆ ADE

tan 60° = Perpendicular / base = DE/AD

√3= h/x [tan 60° = √3]

h = √3x……..(1)

In ∆ ABC

tan 30° = AB /BC

[ tan30° = 1/√3]

1/√3 = 10/x

x= 10√3 m.. …………..(2)

Substitute the value of x from equation (2) in equation (1), we have

h = √3x

h= √3× 10√3= 10 × 3= 30 m

h = 30 m

The height of the hill is CE= CD+ DE= 10 +30= 40 m

Hence, the height of the hill is 40 m & the Distance of the hill from the ship is 10√3 m.

Read more on Brainly.in – https://brainly.in/question/1001119#readmore

Workspace

A.

32 kmph

B.

36 kmph

C.

38 kmph

D.

40 kmph

Answer with explanation

Answer: Option AExplanation

Workspace

From a point P on a level ground, the angle of elevation of the top tower is 30°. If the tower is 100 m high, the distance of point P from the foot of the tower is:

A.

149 m

B.

156 m

C.

173 m

D.

200 m

Answer with explanation

Answer: Option CExplanation

Let AB be the tower.

Then, APB = 30° and AB = 100 m.

AB | = tan 30° = | 1 |

AP | 3 |

AP | = (AB x 3) m |

= 1003 m | |

= (100 x 1.73) m | |

= 173 m. |

Workspace

The angle of elevation of the sun, when the length of the shadow of a tree 3 times the height of the tree, is:

A.

30°

B.

45°

C.

60°

D.

90°

Answer with explanation

Answer: Option AExplanation

Let AB be the tree and AC be its shadow.

Let ACB = .

Then, | AC | = | 3 cot = 3 |

AB |

= 30°.

Workspace

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