Question

The angle of depression, from the top of a light house of two boats are ${45}^o$ and ${30}^o$ towards the west. If the boats are $6m$ apart, then the height of the light house is:

• A. $3(\sqrt{3}+1)m$
• B. $(\sqrt{3}+1)m$
• C. $3(\sqrt{3}-1)m$
• D. $(\sqrt{3}-1)m$

Answer 1

The correct option is A $3(\sqrt{3}+1)m$

Let $AB$ be the light house.

$C$ and $D$ are the boats

angle of Depression formed is ${45}^o$ and ${30}^o$

$\therefore$ i.e, $\angle CAE={45}^o$

$\angle DAE={30}^o$.

$\therefore$ $\angle ACB={45}^o$

$\angle ADB={30}^o$.

and also given $CD=6cm$

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

$\tan {30}^o=\frac{AB}{BD}=\frac{AB}{6+BC}$

$\frac{1}{\sqrt{3}}=\frac{AB}{6+BC}$

$6+BC=\sqrt{3}AB$ as $BC=AB$

$6+AB=\sqrt{3}AB$

$AB=\frac{6}{\sqrt{3}-1}=\frac{6(\sqrt{3}+1)}{2}$

$\therefore AB=3(\sqrt{3}+1)$