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A tower subtends angles $θ, 2 θ$ and $3 θ$ at 3 points $A, B, C$ respectively, lying on a horizontal line through the foot of the tower then the ratio $A B / B C$ equals to

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$\frac{\cos 3 θ}{\cos θ}$

$\frac{\sin 3 θ}{\sin θ}$

$\frac{\tan θ}{\tan 3 θ}$

$\frac{\sin θ}{\sin 3 θ}$

answer is A.

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Detailed Solution

$A B = P B - P A$
$B C = P C - P B$

$\tan θ = \frac{h}{P A} \Rightarrow P A = h \cot θ$
$\tan 2 θ = \frac{h}{P B} \Rightarrow P B = h \cot 2 θ$
$\tan 3 θ = \frac{h}{P C} \Rightarrow P C h \cdot \cot 3 θ$

$\frac{A B}{B C} = \frac{h (\cot 2 θ - \cot θ)}{h (\cot 3 θ - \cot 2 θ)} = \frac{\sin 3 θ}{\sin θ}$

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