$ABC$ is a triangular park with $AB = AC = 100 \text{ metres}$. $A$ vertical tower of height $h$ is situated at the mid-point $P$ of $BC$. If the angles of elevation of the top of the tower $Q$ at $A$ and $B$ are $\cot^{-1}(3\sqrt{2})$ and $\csc^{-1}(2\sqrt{2})$ respectively,then the height of the tower (in metres) is

  • A
    $25$
  • B
    $10\sqrt{5}$
  • C
    $\frac{100}{3\sqrt{3}}$
  • D
    $20$

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