Case Study 2: Ladder between two buildings

In right $\triangle OAP$: $\sin 60^\circ = \frac{AP}{OP} \implies \frac{\sqrt{3}}{2} = \frac{12\sqrt{3}}{OP} \implies OP = 24\text{ m}$ Length of the ladder is $24\text{ m}$.

In right $\triangle OCR$, length of ladder $OR = 24\text{ m}$ and $\angle ROC = 45^\circ$. $\cos 45^\circ = \frac{OC}{OR} \implies \frac{1}{\sqrt{2}} = \frac{OC}{24} \implies OC = \frac{24}{\sqrt{2}} = 12\sqrt{2}\text{ m} \approx 16.97\text{ m}$.

In $\triangle OAP$: $\cos 60^\circ = \frac{OA}{OP} \implies \frac{1}{2} = \frac{OA}{24} \implies OA = 12\text{ m}$ Distance between two buildings $= OC + OA = 12\sqrt{2} + 12 = 12(\sqrt{2} + 1)\text{ m}$.

In right $\triangle OCR$: $\sin 45^\circ = \frac{RC}{OR} \implies \frac{1}{\sqrt{2}} = \frac{RC}{24} \implies RC = 12\sqrt{2}\text{ m}$ Height of building X is $12\sqrt{2}\text{ m}$.