• question_answer 28) As observed from the top of a 100 m high light house from the sea-level, the angles of depression of two ships are $30{}^\circ$ and$45{}^\circ$. If one ship is exactly behind the other on the same side of the light house, find the distance between the two ships. [Use$\sqrt{3}=1.732$]

 Let AB be the light house and two ships be at C and D. In $\Delta \text{ }ABC$, $\frac{BC}{AB}=\cot \,45{}^\circ$ $\Rightarrow$               $\frac{x}{100}=1$ $\Rightarrow$               $x=100$                                              ...(i) Similarly, in $\Delta \text{ }ABD$, $\frac{BD}{AB}=\cot \,\,30{}^\circ$ $\Rightarrow$               $\frac{y}{100}=\sqrt{3}$ $\Rightarrow$               $y=100\sqrt{3}$                                               ?(ii) Distance between two ships $=y-x$ $=100\sqrt{3}-100$ [from (i) and (ii)] $=100\left( \sqrt{3}-1 \right)$ $=100(1.732-1)$ $=73.2\,\,m$