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At a point on level ground, the angle of elevation of a vertical tower is found to be such that its tangent is $$\frac{5}{12}$$. On walking 192 m towards the tower, the tangent of the angle of elevation is $$\frac{3}{4}$$. The height of the tower is

tanα = 5/12
tanβ = 3/4
In triangle BAC,
tanα =AB/AC = 5/12 = h/(X+192)..................(1)
In Triangle DAB,
tanβ = AB/AD = 3/4 = h/x
x = 4h/3
Using (ii) in (i)
5/12 = h/(192 + 4h/3)
On solving, we get
h = 180 metres.
Hence, the height of the tower is 180 metres.
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