Math, asked by electro11233, 4 months ago

a 20m high vertical pole and a vertical tower are on the same level ground in such a way that the angle of elevation of the top of the tower as seen from the foot of the pole is 60° and the angle of elevation of the top of the pole as seen from the foot of the tower is 30° find i )the height of the tower , ii)the horizontal distance between pole and tower ​

Answers

Answered by TheBrainliestUser
71

Answer:

i) Height of tower = 60 m

ii) Horizontal distance between pole and tower = 20√3 m

Step-by-step explanation:

Given: Height of pole = 20 m

Let us suppose that ABC and DBC are two triangles.

So, In △ABC: AB is the height of pole

And BC is horizontal distance between pole and tower.

In △DBC: DC is height of tower.

We have:

The angle of elevation of the top of the tower as seen from the foot of the pole is 60°:

It means in △DBC, ∠DBC = 60°

The angle of elevation of the top of the pole as seen from the foot of the tower is 30°:

It means in △ABC, ∠ACB = 30°

To find:

i) The height of the tower.

ii) The horizontal distance between pole and tower.

Now, In △ABC:

tan30° = AB/BC

→ 1/√3 = 20/BC

→ BC = 20√3

BC is horizontal distance between pole and tower = 20√3 m

In △DBC:

tan60° = DC/BC

→ √3 = DC/20√3

→ DC = 20√3 × √3

→ DC = 60

DC is height of tower = 60 m

Attachments:

rsagnik437: Excellent!
Answered by ItzIshan
51

QuestioN :-

A 20m High vertical pole and a vertical tower are on the same level ground in such a way that the angle of elevation of the top of the tower as seen from the foot of the pole is 60° and the angle of elevation of the top of the pole as seen from the foot of the tower is 30°. Find :-

  • The hight of the tower

  • The horizontal distance between pole and tower.

Given :-

  • Height of pole = 20 m

AnsweR :-

Let , ABC and PQR are two triangles.

So according to the Question , in ABC ,

  • AB = height of pole

  • BC = Horizontal distance between pole and tower.

And in PQR ,

  • PR = height of the tower

In the Question Given that ,

  • Pole and tower are on the level in such way that the angle of elevation of the top of the tower as seen from the foot of the pole = 60°.

So, in angle PQR , PQR = 60°

And ,

  • The angle of elevation of the top of the pole as seen from the foot of the tower is 30°.

So in ABC , ACB = 30°

 \sf in \:   \triangle \:  abc \:  \mapsto \\  \\  \implies \sf \:  \tan(30 \degree)  =  \frac{ab}{bc}  \\  \\  \sf \: but \: ab = 20(given) \: and \: tan(30 \degree) =  \frac{1}{ \sqrt{3} }  \\  \\   \implies \: \sf  \frac{1}{ \sqrt{3} }  =  \frac{20}{bc}  \\  \\  \implies \boxed{ \sf \: bc = 20 \sqrt{3} } \\  \\  \sf \: hence \: horizontal \: distance \: between \:  \\  \sf pole \: and \: tower \: is \: 20 \sqrt{3} .

  \sf \: in \:  \triangle \: pqr  \:  \mapsto  \\  \\   \implies \: \sf \:  \tan(60 \degree)  =  \frac{pr}{bc}  \\  \\   \sf \: but \: bc = 20 \sqrt{3} \: and \:  \tan(60 \degree)   =  \sqrt{3}  \\  \\  \implies \: \sf \:   \sqrt{3}   =  \frac{pr}{20 \sqrt{3} }  \\  \\  \implies \sf \: pr = 20 \sqrt{3}  \times  \sqrt{3}  \\  \\  \implies \sf \: pr = 20 \times 3 \\  \\  \implies  \boxed{\sf pr = 60} \\  \\  \sf \: hence \: the \: height \: of \: the \: tower \: is \: 60 \: m

__________________________

Hope it will help you :)


rsagnik437: Amazing!
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