The angle of elevation of the tower from certain point is 300. If the observer moves 20 m towards the tower, the angle of elevation of the top increase by 150.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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