From a point on the ground, the angles of elevation of the bottom and the top
of a transmission tower fixed at the top of a 20 m high building are 45° and
60° respectively. Find the height of the tower.
[ Class 10 ]
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Answers
Given :
The angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively.
To find :
The height of the tower.
Solution :
Let DC be the height of tower and BC be the height of building.
- Tower, BC = 20m.
- ∠CAB = 45°
- ∠DAB = 60°
Let the height of tower (DC) = ‘h’ m.
So, In right angle triangle ∆ABC,
⟹ tan 45° = BC/AB.
• The value of tan 45° = 1.
⟹ 1 = 20/AB
⟹ AB = 20m.
Now,
In right angle triangle ∆ABD,
⟹ tan 60° = BD/AD
• The value of tan 60° = √3.
⟹ √3 = h+20/20
⟹ h = 20 (√3 - 1).
The height of the tower is 20 (√3 - 1).
Answer:
ANSWER
Let DC be the tower and BC be the building. Then,
∠CAB=45
o
,∠DAB=60
o
,BC=20 m
Let height of the tower, DC=h m.
In right △ABC,
tan45
o
=
AB
BC
1=
AB
20
AB=20 m
In right △ABD,
tan60
o
=
AB
BD
3
=
20
h+20
h=20(
3
−1) m
Explanation:
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