A tower subtends angles $\alpha, 2 \alpha$ and $3 \alpha$ respectively at points $A, B$ and $C$, all lying on a horizontal line through the foot of the tower, then $\frac{A B}{B C}$ is equal to:

Question

Aicharya · Accepted Answer

In $	riangle E C D$, $	an 3 \alpha=\frac{h}{C D}$ $\Rightarrow \quad C D=h \cot 3 \alpha \quad \ldots(\mathrm{i})$ In $	riangle E B D$, $	an 2 \alpha=\frac{h}{B D}$ $\Rightarrow \quad B D=h \cot 2 \alpha \quad \ldots(\mathrm{ii})$ In $	riangle E A D$, $	an \alpha=\frac{h}{A D}$ $\Rightarrow \quad A D=h \cot \alpha \quad \ldots(\mathrm{iii})$ From Eqs. (ii) and (iii), $A D-B D=h \cot \alpha-h \cot 2 \alpha$ $A B=h(\cot \alpha-\cot 2 \alpha) \quad \ldots(\mathrm{iv})$ From Eqs. (i) and (ii), $B D-C D=h \cot 2 \alpha-h \cot 3 \alpha$ $B C=h(\cot 2 \alpha-\cot 3 \alpha) \quad \ldots(\mathrm{v})$ From Eqs. (iv) and (v), $\frac{A B}{B C}=\frac{h(\cot \alpha-\cot 2 \alpha)}{2(\cot 2 \alpha-\cot 3 \alpha)}$ $\Rightarrow \frac{A B}{B C}=\frac{\frac{\cos \alpha}{\sin \alpha}-\frac{\cos 2 \alpha}{\sin 2 \alpha}}{\frac{\cos 2 \alpha}{\sin 2 \alpha}-\frac{\cos 3 \alpha}{\sin 3 \alpha}}=\frac{\frac{\sin (2 \alpha-\alpha)}{\sin \alpha \sin 2 \alpha}}{\frac{\sin (3 \alpha-2 \alpha)}{\sin 2 \alpha \sin 3 \alpha}}$ $=\frac{\sin 3 \alpha}{\sin \alpha}=\frac{3 \sin \alpha-4 \sin ^3 \alpha}{\sin \alpha}$ $=3-4 \sin ^2 \alpha$ $=-3-2(1-\cos 2 \alpha)$ $=1+2 \cos 2 \alpha$

A tower subtends angles $\alpha, 2 \alpha$ and $3 \alpha$ respectively at points $A, B$ and $C$, all lying…

Solution