Q30 of 47 Page 1

An observer finds the angle of elevation of the top of the tower from a certain point on the ground as 30°. If the observer moves 20 m towards the base of the tower, the angle of elevation of the top increases by 15°, find the height of the tower.

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Let height of the tower CD be h m.


Consider ΔDCB,


⇒ tan45° = DC/CB


⇒ 1 = h/CB


⇒ CB = h


Consider ΔDCA,


⇒ tan30° = DC/CA


⇒ 1/√3 = h/CA


Since CA = CB + AB = h + 20


⇒ 1/√3 = h/h + 20


⇒ h + 20 = √3h


⇒ √3h – h = 20


⇒ h (√3 – 1) = 20


⇒ h = 20/ (√3 – 1)


⇒ h = 10 + 10√3


∴ h = 27.32 m


Ans. The height of the tower is 27.32 m.


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