The length of the shadow of a tower standing on level plane is found to be $ 2 x $ metres longer when the sun's altitude is $ 30^{\circ} $ than when it was $ 45^{\circ} $. Prove that the height of tower is $ x(\sqrt{3}+1) $ metres.

Given:

The length of the shadow of a tower standing on level plane is found to be \( 2 x \) metres longer when the sun's altitude is \( 30^{\circ} \) than when it was \( 45^{\circ} \).

To do:

We have to prove that the height of tower is \( x(\sqrt{3}+1) \) metres.

Solution:  

Let $AB$ be the height of the tower, $CA$ be the shadow when Sun's altitude is \( 30^{\circ} \) and $DA$ be the shadow when Sun's altitude is \( 45^{\circ} \).

From the figure,

$\mathrm{CD}=2x \mathrm{~m}, \angle \mathrm{BCA}=30^{\circ}, \angle \mathrm{BDA}=45^{\circ}$

Let the height of the tower be $\mathrm{AB}=h \mathrm{~m}$ and the length of the shadow when Sun's altitude is \( 45^{\circ} \) be $\mathrm{DA}=y \mathrm{~m}$.

This implies,

$\mathrm{CA}=2x+y \mathrm{~m}$

We know that,

$\tan \theta=\frac{\text { Opposite }}{\text { Adjacent }}$

$=\frac{\text { AB }}{DA}$

$\Rightarrow \tan 45^{\circ}=\frac{h}{y}$

$\Rightarrow 1=\frac{h}{y}$

$\Rightarrow h=y \mathrm{~m}$.........(i)

Similarly,

$\tan \theta=\frac{\text { Opposite }}{\text { Adjacent }}$

$=\frac{\text { AB }}{CA}$

$\Rightarrow \tan 30^{\circ}=\frac{h}{2x+y}$

$\Rightarrow \frac{1}{\sqrt3}=\frac{h}{2x+y}$

$\Rightarrow 2x+y=h \sqrt3 \mathrm{~m}$

$\Rightarrow 2x+h=h\sqrt3 \mathrm{~m}$            [From (i)]

$\Rightarrow h(\sqrt3-1)=2x \mathrm{~m}$

$\Rightarrow h=\frac{2x}{\sqrt3-1} \mathrm{~m}$

$\Rightarrow h=\frac{2x(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)} \mathrm{~m}$

$\Rightarrow h=\frac{2x(\sqrt3+1)}{3-1} \mathrm{~m}$

$\Rightarrow h=\frac{2x(\sqrt3+1)}{2} \mathrm{~m}$

$\Rightarrow h=x(\sqrt3+1) \mathrm{~m}$

Hence proved.

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Updated on: 10-Oct-2022

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