The angle of elevation of the top of a vertical tower from a point...

\text{ Let } \mathrm{AB} \text{ be$the$tower of height ' } \mathrm{h} \text{ '. }$\\ Given : In$\triangle \mathrm{ABP}\tan \alpha=\frac{\mathrm{AB}}{\mathrm{PB}}$or$\frac{\sin \alpha}{\cos \alpha}=\frac{h}{x+2}\Rightarrow(x+2) \sin \alpha=\mathrm{h} \cos \alpha\Rightarrow h=\frac{x \sin \alpha+2 \sin \alpha}{\cos \alpha}$Now, In$\Delta \mathrm{ABC}, \tan \beta=\frac{\mathrm{AB}}{\mathrm{BC}} ⇒ \frac{\sin \beta}{\cos \beta}=\frac{h}{x} \Rightarrow$x=$\frac{h \cos \beta}{\sin \beta}$Putting the value of$x$in eq. (2) to eq. (1), we get$$\begin{aligned}&h=\frac{\frac{h \cos \beta \sin \alpha}{\sin \beta}+\frac{2 \sin \alpha}{1}}{\cos \alpha}\\ &\Rightarrow h=\frac{h \cos \beta \cdot \sin \alpha+2 \sin \alpha \sin \beta}{\sin \beta \cdot \cos \alpha}\\ & \end{aligned}$$\Rightarrow h(\sin \beta \cdot \cos \alpha-\cos \beta \cdot \sin \alpha)=2 \sin \alpha \cdot \sin \beta\Rightarrow h[\sin (\beta-\alpha)]=2 \sin \alpha \cdot \sin \beta\Rightarrow h=\frac{2 \sin \alpha \cdot \sin \beta}{\sin (\beta-\alpha)}$