# They call it Height and Distance problem?

**Geometry**Level 3

\(P\) is top and \(Q\) is the foot of a tower standing on a horizontal plane. \(A\) is the foot a building standing beside the tower on the same horizontal plane, \(B\) is another point on the building such that \(AB\) is \(32 \text{ m}\). It is found that,

\[\begin{cases} \cot(\angle PAQ)=\dfrac{2}{5} \\ \cot(\angle PBQ)=\dfrac{3}{5}\end{cases} \]

If the height of tower (in meters) is in the form \(a\sqrt{b}-c\) where \(a,b,c \in \mathbb{N}\) and \(b\) is square-free, then find \(a+b+c\).