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
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
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°
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