sir this is the question from the book -ML aggarwal( avichal publication) class 10th , chapter20 , heights and distances
we have been given that At a point on level ground,
the angle of elevation of a vertical lower is found to be such that its tangent is 5/12.
On walking 192 m towards the tower, the tangent of the angle is found to be ¾.
wehave to Find the height of the tower.
question no 21 , heights and distances , ICSE board, ML Aggarwal
Consider TR as the tower and P as the point on the ground such that tan θ = 5/12
tan α = ¾
PQ = 192 m
Take TR = x and QR = y
In right triangle TQR
tan α = TR/QR = x\y
So we get
3/4 = x/y
y = 4/3 x …..(1)
In right triangle TPR
tan θ = TR/PR
Substituting the values
5/12 = x/ (y + 192)
x = (y + 192) 5/12 …… (2)
Using both the equations
x = (4/3 x + 192) 5/12
So we get
x = 5/9x + 80
x – 5/9 x = 80
4/9 x = 80
By further calculation
x = (80 × 9)/ 4 = 180
Hence, the height of the tower is 180 m.