The angle of elevation of the top of a tower from a point A (on the ground) is 300. On walking 50 m towards the tower, the angle of elevation is found to be 600. Calculate
(i) the height of the tower (correct to one decimal place).
(ii) the distance of the tower from A.
Solution:
Consider TR as the tower and A as the point on the ground
Angle of elevation of the top of tower = 300
AB = 50 m
Angle of elevation from B = 600
Take TR = h and AR = x
BR = x – 50
In right triangle ATR
tan θ = TR/AR
Substituting the values
tan 300 = h/x
So we get
1/√3 = h/x
x = √3h ……. (1)
In right triangle BTR
tan θ = TR/BR
Substituting the values
tan 600 = h/ (x – 50)
So we get
√3 = h/ (x – 50)
h = √3 (x – 50) …… (2)
Using both the equations
h = √3 (√3h – 50)
By further calculation
h = 3h – 50√3
2h = 50√3
h = 25 √3
So we get
h = 25 × 1.732 = 43.3
Now substituting the values of h in equation (1)
x = √3 × 25√3
x = 25 × 3
x = 75
Here
Height of the tower = 43.3 m
Distance of A from the foot of the tower = 75 m