Question

The angles of depression, from the top of a light house, of two boats are ${45}^{\circ }$ and ${30}^{\circ }$ towards the west. If the two boats are 6 m 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 the height of the light house $AB=h$

Then in $△ABC,\tan {45}^{\circ }=\frac{AB}{BC}$

$\Rightarrow AB=BC$ since $\tan {45}^{\circ }=1$

Now in $△ABD,\tan {30}^{\circ }=\frac{h}{DC+CB}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{DC+CB}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{DC+CB}=\frac{AB}{DB}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{h+6}$

$\Rightarrow h\sqrt{3}=h+6$

$\Rightarrow h(\sqrt{3}-1)=6$

$\Rightarrow h=\frac{6}{(\sqrt{3}-1)}$ by rationalising the denominator we get

$\Rightarrow h=\frac{6}{(\sqrt{3}-1)}\times \frac{(\sqrt{3}+1)}{(\sqrt{3}+1)}=\frac{6(\sqrt{3}+1)}{3-1}=3(\sqrt{3}+1)$