Let AB be the light house C and D are two ships whose angles of depression on A are 30° and 45° respectively Let height of the tower = h $\begin{array}{l}\angle\mathrm{ACB}=\angle\mathrm{XAC}=30^{\circ},\\ \angle\mathrm{ADB}=\angle\mathrm{YAD}=45^{\circ}\\ \text{ and }\mathrm{CD}=100\mathrm{~m}\\ \text{ Let }\mathrm{AB}=\mathrm{h}\text{ and }\mathrm{CB}=\mathrm{x}\\ \text{ then }\mathrm{BC}=(100-\mathrm{x})\mathrm{m}\\ \text{ Now in }\Delta\text{ACB},\\ \tan\theta=\frac{\text{AB}}{\text{CB}}\\ \tan30^{\circ}=\frac{\text{h}}{\text{x}}\\ \Rightarrow\frac{1}{\sqrt{3}}=\frac{\text{h}}{\text{x}}\\ \Rightarrow x=\sqrt{3}\text{h}\ldots.(\mathrm{i})\\ \text{Similarily in }\triangle\text{ADB}\\ \tan45^{\circ}=\frac{\text{AB}}{\text{BD}}\\ \Rightarrow1=\frac{\text{h}}{100-\text{x}}\\ \Rightarrow x=100-\text{h}\ldots\text{ (ii) }\\ \text{ From (i) and (ii) }\\ \sqrt{3}\text{h}=100-\text{h}\\ \Rightarrow(\sqrt{3}-1)\text{h}=100\\ \text{h}=\frac{100}{\sqrt{3}+1}\\ \quad=\frac{100(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}\\ \quad=\frac{100(\sqrt{3}-1)}{3-1}\\ \quad=\frac{100(\sqrt{3}-1)}{2}\\ \quad=50(\sqrt{3}-1)\\ \therefore\text{ height of light house }\\ =50(\sqrt{3}-1)\ \text{m}\end{array}$