A ladder set against a wall at an angle $\displaystyle 45$° to the ground. If the foot of the ladder is pulled away from the wall through a distance of $\displaystyle 4$ m, its top slides a distance of $\displaystyle 3$ m down the wall making an angle $\displaystyle 30$° with the ground. Find the final height of the top of the ladder from the ground and length of the ladder.

\[\begin{aligned} & \sin 45^{\circ}=\frac{A B}{B D}=\frac{h+3}{B D} \\ & \Rightarrow \mathrm{BD}=(\mathrm{h}+3) \sqrt{2} \end{aligned} \] \[\sin 30^{\circ}=\frac{1}{2}=\frac{h}{C E} \] \[\Rightarrow C E=2 h \] length of ladder remains same Therefore \(\displaystyle \mathrm{BD}=\mathrm{CE} \Rightarrow(\mathrm{h}+3) \sqrt{2}=2 \mathrm{~h}\) \(\displaystyle \Rightarrow \mathrm{h}=\frac{3 \sqrt{2}}{2-\sqrt{2}}=3(\sqrt{2}+1)\) Final height of the top of the ladder \(\displaystyle =3(\sqrt{2}+1) \mathrm{m}\) and length of ladder \(\displaystyle =2 \mathrm{~h}=6(\sqrt{2}+1) \mathrm{m}\)