Question 69

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

Solution

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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