A tower subtends angles alpha, 2 alpha and 3 | vedclass

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Question

(B) In $	riangle ECD$, $	an 3 \alpha = \frac{h}{CD} \Rightarrow CD = h \cot 3 \alpha \quad \dots(i)$In $	riangle EBD$, $	an 2 \alpha = \frac{h}{BD

Vedclass · Accepted Answer

$1+2 \cos 2 \alpha$

Vedclass · Answer

(B) In $	riangle ECD$, $	an 3 \alpha = \frac{h}{CD} \Rightarrow CD = h \cot 3 \alpha \quad \dots(i)$In $	riangle EBD$, $	an 2 \alpha = \frac{h}{BD} \Rightarrow BD = h \cot 2 \alpha \quad \dots(ii)$In $	riangle EAD$, $	an \alpha = \frac{h}{AD} \Rightarrow AD = h \cot \alpha \quad \dots(iii)$From Eqs. $(ii)$ and $(iii)$, $AB = AD - BD = h(\cot \alpha - \cot 2 \alpha) \quad \dots(iv)$From Eqs. $(i)$ and $(ii)$, $BC = BD - CD = h(\cot 2 \alpha - \cot 3 \alpha) \quad \dots(v)$Dividing $(iv)$ by $(v)$:$\frac{AB}{BC} = \frac{\cot \alpha - \cot 2 \alpha}{\cot 2 \alpha - \cot 3 \alpha} = \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}}$$= \frac{\frac{\sin(2 \alpha - \alpha)}{\sin \alpha \sin 2 \alpha}}{\frac{\sin(3 \alpha - 2 \alpha)}{\sin 2 \alpha \sin 3 \alpha}} = \frac{\sin \alpha}{\sin \alpha \sin 2 \alpha} 	imes \frac{\sin 2 \alpha \sin 3 \alpha}{\sin \alpha} = \frac{\sin 3 \alpha}{\sin \alpha}$$= \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 on a horizontal line through the foot of the tower. Then $\frac{A B}{B C}$ is equal to:

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