The length of the shadow of a tower standing on level ground is found to be 2x metres longer when the sun's elevation is 30° than when it was 45°. The height of the tower is


In the above figure, let AB be the tower and BC and BD are the consecutive shadows of AB for two different positions of the sun.


From ∆DBA,



1 = h/y


h = y


or,


DB = BA


Now, from ∆ACB,



or,



On cross multiplying we get,


2x = √3(h+y)


2x + y = √3


2x + h = (√3h)


2x = (√3 – 1)h


x = (√3 – 1)h



h = (√3 + 1)x

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