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

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(B) Let the tower be $ED$ with height $h$ and $D$ be the foot of the tower. Let $CD = y$. In $	riangle ECD$,$	an 3\alpha = h/y$,so $y = h \cot 3\alp

Vedclass · Accepted Answer

$1 + 2\cos 2\alpha$

Vedclass · Answer

(B) Let the tower be $ED$ with height $h$ and $D$ be the foot of the tower. Let $CD = y$. In $	riangle ECD$,$	an 3\alpha = h/y$,so $y = h \cot 3\alpha$.In $	riangle EBD$,$	an 2\alpha = h/(y+BC)$,so $y+BC = h \cot 2\alpha$,which gives $BC = h(\cot 2\alpha - \cot 3\alpha) = h \frac{\sin(3\alpha - 2\alpha)}{\sin 2\alpha \sin 3\alpha} = \frac{h \sin \alpha}{\sin 2\alpha \sin 3\alpha}$.In $	riangle EAD$,$	an \alpha = h/(y+BC+AB)$,so $y+BC+AB = h \cot \alpha$,which gives $AB = h(\cot \alpha - \cot 2\alpha) = h \frac{\sin(2\alpha - \alpha)}{\sin \alpha \sin 2\alpha} = \frac{h}{\sin 2\alpha}$.Thus,$AB/BC = \frac{h}{\sin 2\alpha} 	imes \frac{\sin 2\alpha \sin 3\alpha}{h \sin \alpha} = \frac{\sin 3\alpha}{\sin \alpha} = \frac{3\sin \alpha - 4\sin^3 \alpha}{\sin \alpha} = 3 - 4\sin^2 \alpha = 3 - 2(1 - \cos 2\alpha) = 1 + 2\cos 2\alpha$.

$A$ tower subtends angles $\alpha, 2\alpha, 3\alpha$ respectively at points $A, B$ and $C$,all lying on a horizontal line through the foot of the tower. Then $AB/BC = $

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