The shadow of a tower standing on a level plane is found to be 50 m longer when Sun’s elevation is 30° than when it is 60°. Find the height of the tower
Solution:
Given, the shadow of a tower standing on a level plane is found to be 50 m longer when the sun's elevation is 30° than when it is 60°
We have to find the height of the tower.
Let SQ be the height of the tower.
SQ = h m
Angle of elevation at first, ∠SRQ = 60°
At first, the length of the shadow = x m
The shadow of the tower is increased by 50 m when the angle of elevation ∠SPQ = 30°
In triangle SRQ,
tan 60° = SQ/RQ
√3 = h/x
x = h/√3 m --------------- (1)
In triangle SPQ,
tan 30° = SQ/PQ
We know that, PQ = PR + RQ
PQ = 50 + x
1/√3 = h/(50+x)
50 + x = √3h
x = √3h - 50 ------------ (2)
From (1) and (2),
h/√3 = √3h - 50
By grouping,
h/√3 - √3h = -50
√3h - h/√3 = 50
On simplification,
(3h-h)/√3 = 50
2h/√3 = 50
2h = 50√3
h = 50√3/2
h = 25√3 m
Therefore, the height of the tower is 25√3 m.
✦ Try This: The shadow of a tower standing on a level plane is found to be 40 m longer when Sun’s elevation is 30° than when it is 45°. Find the height of the tower.
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 8
NCERT Exemplar Class 10 Maths Exercise 8.4 Problem 7
The shadow of a tower standing on a level plane is found to be 50 m longer when Sun’s elevation is 30° than when it is 60°. Find the height of the tower
Summary:
The shadow of a tower standing on a level plane is found to be 50 m longer when Sun’s elevation is 30° than when it is 60°. The height of the tower is 25√3 m
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