Q24 of 46 Page 1

The shadow of a tower standing on a level ground is found to be 20 m longer when the Sun’s altitude is 45o than when it is 60o. Find the height of the tower.

We have


Here, we are given that CD = 20 m, CB = x m, AB = h m, ∠ADB = 45° and ∠ACB = 60°


In ∆ACB,


[∵, ]


⇒ √3 = h/x [∵, tan 60° = √3]


⇒ x√3 = h


Or h = x√3 …(i)


In ∆ADB,


[∵, ]


⇒ [∵, tan 45° = 1]


⇒ 20 + x = h


Or h = x + 20 …(ii)


Substituting equation (i) in equation (ii), we get


x√3 = x + 20


⇒ x√3 – x = 20


⇒ (√3 – 1)x = 20


⇒


Rationalizing it,



⇒ [∵, (a + b)(a – b) = a2 – b2, and so (√3 + 1)(√3 – 1) = 3 – 1)


⇒ .


⇒ x = 10(√3 + 1) = 10(1.73 + 1) = 10 × 2.73


⇒ x = 27.3


Putting x = 27.3 in equation (ii), we get


h = 27.3 + 20


⇒ h = 47.3


Thus, the height of the tower is 47.3 m.


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