Question

From the top of a lighthouse $70m$ high with its base at sea level, the angle of depression of a boat is ${15}^o$. The distance of the boat from the foot of the lighthouse is:

• A. $70(2-\sqrt{3})m$
• B. $70(2+\sqrt{3})m$
• C. $70(3-\sqrt{3})m$
• D. $70(3+\sqrt{3})m$

Answer 1

The correct option is A $70(2-\sqrt{3})m$

Let AB be the light house and the boat is at point C
Angle BCA is also equal to angle of depression $={15}^o$

$\Rightarrow \angle BCA={15}^o$

Now, $\tan \left({15}^o\right)=\frac{AB}{BC}$

$\Rightarrow BC=AB\left(\tan \right({15}^o\left)\right)\tan \left({15}^o\right)=\tan ({60}^o-{45}^o)=\frac{\tan {60}^o-\tan {45}^o}{1+\tan {60}^o\tan {45}^o}=\frac{\sqrt{3}-1}{1+\sqrt{3}}=\frac{{(\sqrt{3}-1)}^2}{3-1}=2-\sqrt{3}$

So, $BC=70(2-\sqrt{3})m$

Option A is correct