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 tangents of the angle is found to be 3/4. Find the height of the tower.
Solution
Let AB be the tower of height lim.D and C are points on the level ground such that distance between them are 192 mts. Let the angles of elevation of a vertical tower from points D and C be α and β respectively.
i.e., ∠ACB = β
and ∠ADB = α
Let BC = x metres
In right triangle ABC, we have
In right triangle ABD, we have
Comparing (i) and (ii), we get
Hence, the height of the tower be 180 mts.
