The angle of elevation of the tower from certain point is 30^{0}. If the observer moves 20 m towards the tower, the angle of elevation of the top increase by 15^{0}.Find the height of the tower.

Let PR = h meter, be the height of the tower.

The observer is standing at point Q such that, the distance between the observer and tower is QR = (20+x) m, where

QR = QS + SR = 20 + x

∠PQR = 30°

∠ PSR = θ

In ∆PQR,

Rearranging the terms,

We get 20 +x = √3 h

In ∆PSR,

Since, angle of elevation increases by 15 when the observer moves 20 m towards the tower. We have,

θ = 30° + 15° = 45°

So,

⇒ h = x

Substituting x=h in eq. 1, we get

h = √3 h – 20

⇒ √3 h – h = 20

⇒ h (√3 - 1) = 20

Rationalizing the denominator, we have

= 10 (√3 + 1)

Hence, the required height of the tower is 10 (√3 + 1) meter.

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