The angle of elevation of the top of a vertical tower from a point on the ground is 60^{0}.From another point 10 m vertically above the first its angle of elevation is 45^{0}. Find the height of the tower.

Given that ∠TQO = 60° and ∠TAB = 45°, when AP = 10 m.

So, OB = 10 m from the diagram.

Let PO = x m & if height of the vertical tower, TO = H m, then TB = TO –OB = (H -10) m

In ∆TPO,

In ∆TAB,

⇒ AB = H -10

[∵ TO –TH = H-10]

⇒ x = H -10 …eq. 2

Comparing eq. 1 and eq. 2, we get

Rationalizing it,

Hence, required height of the tower is 5√3 (√3+1) m.

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