The shadow of a tower standing on a level plane is found

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The shadow of a tower standing on a level plane is found to be 30 metre longer when the Sun’s altitude changes from 60° to 45°. The height of the tower is

1. 15 (3 + √3 ) metre
2. 15 ( √3 + 1) metre
3. 15 ( √3 – 1) metre
4. 15 (3 – √3 ) metre

AB = Tower = h metre (let)
CD = 30 metre
BC = x metre (let)
From ∆ABC,

tan 60° =	AB
	BC

⇒ √3 =	h
	x

⇒ h = √3x metre ...........(i)
From ∆ABD,

tan 45° =	AB
	BD

⇒ 1 =	h
	x + 30

⇒ h = x + 30

⇒ h=	h	+ 30
	√3

⇒ √3h = h + 30 √3
⇒ √3h – h = 30 √3
⇒ h ( √3 - 1) = 30 √3

⇒ h =	30√3	=	30√3(√3 + 1)
	√3 - 1		(√3 - 1)(√3 + 1)

=	30√3(√3 + 1)
	3 - 1

= 15 √3 ( √3 +1)
= 15 (3 + √3 ) metre