Question

The shadow of a tower, when the angle of elevation of the sun is ${45}^o$, is found to be 10 metres longer than when the angle of elevation is ${60}^o$. Find the height of the tower.

• A. $15+5\sqrt{3}m$
• B. $12+5\sqrt{3}m$
• C. $15+\sqrt{3}m$
• D. $12+\sqrt{3}m$

Answer 1

The correct option is A $15+5\sqrt{3}m$

Let AB be the tower and let AC and AD be its shadows when the angles of elevation of the sum are ${60}^o$ and ${45}^o$, respectively. Then,

$\angle ACB={60}^o,\angle ADB={45}^o,CD=10m$

Let $AB=hmetres$ and $AC=xmetres$

In right $△$CAB, we have,

$\tan {60}^o=\frac{AB}{AC}$

$\sqrt{3}=\frac{h}{x}$

$x=\frac{h}{\sqrt{3}}$ ......(1)

In right $△$ DAB, we have,

$\tan {45}^o=\frac{AB}{AD}$

$1=\frac{h}{10+x}$

$x=h-10$ .....(2)

From 1 and 2, we get,

$\frac{h}{\sqrt{3}}=h-10$

$h=\sqrt{3}h-10\sqrt{3}$

$\sqrt{3}h-h=10\sqrt{3}$

$h(\sqrt{3}-1)=10\sqrt{3}$

$h=5\sqrt{3}(\sqrt{3}+1)$

$h=15+5\sqrt{3}m$