The shadow of a tower standing on a level plane is found to be $50 m$ longer when sun's elevation is $30^{\circ}$ than when it is $60^{\circ} .$ Find the height of the tower.

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Solution

$\begin{matrix} L e t A B b e t o w e r B C b e o r i g i n a l s h a d o w . B D b e f i n a l s h a d o w . G i v e n B D - B C = C D = 50 m I n Δ A B C, tan 60^{\circ} = \frac{A B}{C D} \Rightarrow \frac{A B}{\sqrt{3}} = B C - - - - (1) I n Δ A B D tan 30^{o} = \frac{A B}{B D} \Rightarrow \frac{1}{\sqrt{3}} = \frac{A B}{(B C + C D)} \Rightarrow A B \sqrt{3} = \frac{A B}{\sqrt{3}} + 50 f r o m (1) \Rightarrow A B \sqrt{3} - \frac{A B}{\sqrt{3}} = 50 \Rightarrow 3 A B - \frac{A B}{\sqrt{3}} = 50 ∴ A B = 50 \times \frac{\sqrt{3}}{2} = 25 \times 1.735 = 43.30 m W h i c h i s t h e r e q u i r e d a n s w e r . \end{matrix}$

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The shadow of a tower standing on a level plane is found to be 50m longer when sun's elevation is 30∘ than when it is 60∘. Find the height of the tower.

LetABbetowerBCbeoriginalshadow.BDbefinalshadow.GivenBD−BC=CD=50 mInΔABC,tan60∘=ABCD⇒AB√3=BC−−−−(1)InΔABDtan30o=ABBD⇒1√3=AB(BC+CD)⇒AB√3=AB√3+50from(1)⇒AB√3−AB√3=50⇒3AB−AB√3=50∴AB=50×√32=25×1.735=43.30 mWhichistherequiredanswer.

The shadow of a tower standing on a level plane is found to be $50 m$ longer when sun's elevation is $30^{\circ}$ than when it is $60^{\circ} .$ Find the height of the tower.