The angle of elevation of the top of a tower from a point A on the ground is 30°. On moving a distance of 20 metres towards the foot of the tower to a point B the angle of elevation increases to 60°. Find the height of the tower and the distance of the tower from the point A.
Answers
Answer:
The height of the tower is 173.2 m & the distance of the tower from the point A is 30 m.
Step-by-step explanation:
Let CD be the height of the tower (h).
Given :
Angle of elevation of the top of the tower (θ), ∠CDA = 30°
On moving a distance AB = 20 m towards the foot of the tower,then the Angle of elevation of the top of the tower ,(θ), ∠DBC = 60°
Let CB = x m
In right angle triangle, ∆DCB,
tan θ = P/ B
tan 60° = CD/BC
√3 = h/x
x = h/√3 ……...…..(1)
In right angle triangle, ∆DAC
tan θ = P/ B
tan 30° = CD/AC
1/√3 = CD/(BC + BA)
1/√3 = h/(x + 20)
√3h = x + 20
√3h = h/√3 + 20
[From eq 1]
√3h - h/√3 = 20
(√3 ×√3 h - h)/√3 = 20
(3h - h)/√3 = 20
2h = 20√3
h = (20√3)/2
h = 10√3
h = 10 × 1.732
[√3 = 1.732]
h = 17.32 m
FOR DISTANCE :
x = h/√3
x = 17.32/1.732
x = 10 m
AC = x + 20
AC = 10 + 20
AC = 30 m
Hence, the height of the tower is 173.2 m & the distance of the tower from the point A is 30 m.
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Solution
Let
The height of a tower is CD .
Distance between AC is xm
The distance between point A and B is 20 m
BC = x -20m
Now , In∆ BCD
tan 60° = CD/ BC
√3. = h/ x - 20
x -20. = h/√3. eq (I)
In ∆ ACD
tan 30° = CD/ AC
1/√3. = h/ x
x. = √3 h eq 2nd
Putting the value of x in eq 1 st
√3 h - 20 = h/√3
√3(√3h - 20 ) = h
3h - 20√3. = h
3h - h. = 20√3
2h. = 20√3
h. = 20√3/2= 10√3 m
So , height of tower (DC ) is 10√3 m.
Putting the value of h in eq 2nd
√3 h = x
√3 (10√3) = x
30. m. = x
So ,( x) AC is 30 m
Hence , height of tower is 10√3 m and distance between point A and foot of tower is 30 m .
