A tower substends angles $\mathrm{\alpha },2\mathrm{\alpha },3\mathrm{\alpha }$ at A,B, C all lying on the horizontal line through the foot of the tower then $\frac{\mathrm{AB}}{\mathrm{BC}}=$

$\begin{array}{l}\cot \mathrm{\alpha }=\frac{\mathrm{a}+\mathrm{b}+\mathrm{x}}{\mathrm{h}}\to \left(1\right)\\ \cot 2\mathrm{\alpha }=\frac{\mathrm{b}+\mathrm{x}}{\mathrm{h}}\to \left(2\right)\\ \cot 3\mathrm{\alpha }=\frac{\mathrm{x}}{\mathrm{h}}\to \left(3\right)\\ Now \left(1\right)-\left(2\right)\Rightarrow a=\mathrm{h}(\cot \mathrm{\alpha }-\cot 2\mathrm{\alpha })\\ \left(2\right)-3)\Rightarrow b=\mathrm{h}(\cot 2\mathrm{\alpha }-\cot 3\mathrm{\alpha })\\ \frac{\mathrm{a}}{\mathrm{b}}=\frac{\mathrm{h}(\cot \mathrm{\alpha }-\cot 2\mathrm{\alpha })}{\mathrm{h}(\cot 2\mathrm{\alpha }-\cot 3\mathrm{\alpha })}\\ =1+2\cos 2\mathrm{\alpha }\end{array}$