The length of the shadow of a tower standing on level plane is found to be 2x

Question:

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.

Solution:

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 $x(\sqrt{3}+1) \mathrm{m}$.