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(B) Let the height of the tower $ED = h$. In $	riangle EDC$,$	an(60^o) = \frac{ED}{CD} \implies \sqrt{3} = \frac{h}{CD} \implies CD = \frac{h}{\sqrt

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$\sqrt{3} : 1$

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(B) Let the height of the tower $ED = h$. In $	riangle EDC$,$	an(60^o) = \frac{ED}{CD} \implies \sqrt{3} = \frac{h}{CD} \implies CD = \frac{h}{\sqrt{3}}$. In $	riangle EDB$,$	an(45^o) = \frac{ED}{BD} \implies 1 = \frac{h}{BD} \implies BD = h$. In $	riangle EDA$,$	an(30^o) = \frac{ED}{AD} \implies \frac{1}{\sqrt{3}} = \frac{h}{AD} \implies AD = h\sqrt{3}$. Now,$BC = BD - CD = h - \frac{h}{\sqrt{3}} = h\left(\frac{\sqrt{3}-1}{\sqrt{3}}ight)$. $AB = AD - BD = h\sqrt{3} - h = h(\sqrt{3}-1)$. Therefore,the ratio $\frac{AB}{BC} = \frac{h(\sqrt{3}-1)}{h\left(\frac{\sqrt{3}-1}{\sqrt{3}}ight)} = \sqrt{3} : 1$.

If the angles of elevation of the top of a tower from three collinear points $A, B$ and $C$ on a line leading to the foot of the tower are $30^o, 45^o$ and $60^o$ respectively,then the ratio $AB : BC$ is:

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