Q20 of 35 Page 219

At a point on level ground, the angle of elevation of a vertical tower is found to be such that its tangent is . On walking 192 metres towards the tower, the tangent of the angle is found to be. Find the height of the tower.


Let AB be the tower.


The angle of elevation of tower from D is α and on walking 192 metres towards the tower from D to B, the angle of elevation of tower from C is ß.


CD = 192 m


,


Let BC = x, AB = h


BD = BC + CD = x + 192


∠ADB = α ∠ACB = ß


In ∆ABC,




4h = 3x


…..equation (1)


In ∆ABD,




12h = 5x + 960



36h = 20h + 2880


16h = 2880


H = 180


The height of the tower = AB = h = 180 m.


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