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Q.

A tower $P Q$ stands at a point $P$ with in the triangular park $A B C$ such that the sides $a, b, c$ of the triangle subtend equal angles at $P,$ the foot of the tower and the tower subtends angles $α, β, γ$ at $A, B, C$ respectively, then $a^{2} (\cot β - \cot γ) + b^{2} (\cot γ - \cot α) + c^{2} (\cot α - \cot β)$ is equal to

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a

$- 1$

b

0

c

1

d

$a + b + c$

answer is B.

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Detailed Solution

Let $h$ be the height of the tower at $P,$ then

$P A = h \cot α, P B = h \cot β, P C = h \cot γ$

From triangle $P B C$

$\begin{array}{l} \cos 120^{\circ} = \frac{h^{2} \cot^{2} γ + h^{2} \cot^{2} β - a^{2}}{2 h^{2} \cot β \cot γ} \\ \Rightarrow a^{2} = h^{2} [\cot^{2} β + \cot^{2} γ + \cot β \cot γ] \\ \Rightarrow a^{2} (\cot β - \cot γ) = h^{2} [\cot^{3} β - \cot^{3} γ] \end{array}$

Similarly for $b^{2} (\cot γ - \cot α)$ and
$c^{2} (\cot α - \cot β)$

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