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Question
At a point on level ground, the angle of elevation of a vertical tower is found to be such that its tangent is `5/12`. On walking 192 metres towards the tower, the tangent of the angle is found to be `3/4`. Find the height of the tower.
Solution
Let AB be the vertical tower and C and D be two point such that CD = 192 m. Let ∠ACB = θ and ∠ADB = α
Given, `tan theta = 5/12`
`=> (AB)/(BC) = 5/12`
`=> AB = 5/12 BC` ...(i)
Also, `tan alpha = 3/4`
`=> (AB)/(BD) = 3/4`
`=> ((5)/(12)BC)/(BD) = 3/4`
`=> (192 + BD)/(BD) = 3/4 xx 12/5`
`=>` BD = 240 m
∴ BC = (192 + 240) = 432 m
∴ By (i), `AB = 5/12 xx 432 = 180 m`
Hence, the height of the tower is 180 m.
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