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.
Let AB be the tower of height h m. the length of shadow of tower to be found 2x meters at the plane longer when sun’s altitude is 30° than when it was 45°. Let BC = y m,
$C D=2 \times \mathrm{m}$ and $\angle A D B=30^{\circ}, \angle A C B=45^{\circ}$
We have to find the height of the tower
We have the corresponding figure as follows
So we use trigonometric ratios.
In a triangle,
$\Rightarrow \quad \tan C=\frac{A B}{B C}$
$\Rightarrow \quad \tan 45^{\circ}=\frac{h}{v}$
$\Rightarrow \quad 1=\frac{h}{y}$
$\Rightarrow \quad y=h$
Again in a triangle $A D B$
$\Rightarrow \quad \tan D=\frac{A B}{B C+C D}$
$\Rightarrow \quad \tan 30^{\circ}=\frac{h}{2 x+y}$
$\Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{h}{2 x+y}$
$\Rightarrow \quad \sqrt{3} h=2 x+y$
$\Rightarrow \quad \sqrt{3} h=2 x+h$
$\Rightarrow \quad h=\frac{2 x}{(\sqrt{3}-1)}$
$\Rightarrow \quad h=x(\sqrt{3}+1)$
Hence the height of tower is $x(\sqrt{3}+1) \mathrm{m}$.
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