Q. 13

# 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 isA.(2√3x)mB. ( 3√2x) mC. (√3 —1)x mD. (√3 + 1)x m

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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