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

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(B) Let the height of the tower be $h$ and the foot of the tower be $D$. In $	riangle E D C$,$	an 3 \alpha = \frac{h}{C D} \Rightarrow C D = h \cot

Vedclass · Accepted Answer

$1+2 \cos 2 \alpha$

Vedclass · Answer

(B) Let the height of the tower be $h$ and the foot of the tower be $D$. In $	riangle E D C$,$	an 3 \alpha = \frac{h}{C D} \Rightarrow C D = h \cot 3 \alpha$. In $	riangle E D B$,$	an 2 \alpha = \frac{h}{B D} \Rightarrow B D = h \cot 2 \alpha$. In $	riangle E D A$,$	an \alpha = \frac{h}{A D} \Rightarrow A D = h \cot \alpha$. Now,$A B = A D - B D = h(\cot \alpha - \cot 2 \alpha)$ and $B C = B D - C D = h(\cot 2 \alpha - \cot 3 \alpha)$. Therefore,$\frac{A B}{B C} = \frac{\cot \alpha - \cot 2 \alpha}{\cot 2 \alpha - \cot 3 \alpha} = \frac{\frac{\cos \alpha}{\sin \alpha} - \frac{\cos 2 \alpha}{\sin 2 \alpha}}{\frac{\cos 2 \alpha}{\sin 2 \alpha} - \frac{\cos 3 \alpha}{\sin 3 \alpha}} = \frac{\frac{\sin(2 \alpha - \alpha)}{\sin \alpha \sin 2 \alpha}}{\frac{\sin(3 \alpha - 2 \alpha)}{\sin 2 \alpha \sin 3 \alpha}} = \frac{\sin \alpha}{\sin \alpha \sin 2 \alpha} 	imes \frac{\sin 2 \alpha \sin 3 \alpha}{\sin \alpha} = \frac{\sin 3 \alpha}{\sin \alpha}$. Using $\sin 3 \alpha = 3 \sin \alpha - 4 \sin^3 \alpha$,we get $\frac{\sin 3 \alpha}{\sin \alpha} = 3 - 4 \sin^2 \alpha = 3 - 2(1 - \cos 2 \alpha) = 1 + 2 \cos 2 \alpha$.

$A$ 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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