Question

The length of the shadow of a vertical tower on level ground increase by 10 meters when the altitude of the sun changes from ${45}^o$ to ${30}^o$. Then the height of the tower is:

• A. $5\sqrt{3}$
• B. $10(\sqrt{3}+1)meter$
• C. $5(\sqrt{3}+1)$ meter
• D. $10\sqrt{3}meter$

Answer 1

The correct option is C $5(\sqrt{3}+1)$ meter

$Let\text{the distance when the sun elevation is 45 be x}$
$\text{whereas when sun elevation is 30 the distance is 10+x}$
$\text{let h be the height of the tower}$
$\text{when the sun is at elevation of 45}$
$\text{tan45=h/x}$
$\text{1=h/x}$
$\text{h=x}.........\left(i\right)$
$whenthesunelevationis{30}^0$
$\tan {30}^0=h/(x+10)$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x+10}$
$\Rightarrow h=\frac{x+10}{\sqrt{3}}......\left(ii\right)$
$from\left(i\right)and\left(ii\right)$
$x=\frac{x+10}{\sqrt{3}}$
$\Rightarrow \sqrt{3}x=x+10$
$\Rightarrow x(\sqrt{3}-1)=10$
$\Rightarrow x=\frac{10}{\sqrt{3}-1}$
$=\frac{10}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}$
$=5(\sqrt{3}+1)$
$henceheight=5(\sqrt{3}+1)m$