Ex 9.1 Q7 - 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.
Solution:- *CB be the height of the building
*AC be the height of the tower
*D be the point of observation
Given,
*Height of the building = CB = 20 m
*Angle of elevation to the bottom of the tower from point D = ∠CDB = 45°
*Angle of elevation to the top of the tower from point D = ∠ADB = 60°
➙ In ∆CDB, ∠CBD = 90°,
∴ tanD =
Opposite side of ∠D
/
Adjacent side of ∠D
∴ tan45° =
CB
/
DB
∴ tan45° =
20
/
DB
{ Given } ∴ 1 =
20
/
DB
{ ∵ tan45° = 1 } ∴ DB = 20 ------- (i)
➙ In ∆ADB, ∠ABD = 90°,
∴ tanD =
Opposite side of ∠D
/
Adjacent side of ∠D
∴ tan60° =
AB
/
DB
∴ tan60° =
AB
/
20
{ from (i) } ∴ √3 =
AB
/
20
{ ∵ tan60° = √3 } ∴ AB = 20√3
Now,
*Height of the tower
= AC
= AB - CB
= 20√3 - 20
= 20(√3 - 1)
Hence,
The height of the tower is 20(√3 - 1) m.
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Queries Solved:-
Class 10 Ex 9.1
Ex 9.1 Q7 Class 10
Class 10 Ex 9.1 Q7
Class 10 Chap 9 Ex 9.1 Q7
Class 10 Some Applications Of Trigonometry
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