A tower stands vertically inside an acute-ang | vedclass

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(B) Let the height of the tower be $h$ and its distances from the vertices $P, Q,$ and $R$ be $d_1, d_2,$ and $d_3$ respectively.Let the angle of elev

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circumcentre

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(B) Let the height of the tower be $h$ and its distances from the vertices $P, Q,$ and $R$ be $d_1, d_2,$ and $d_3$ respectively.Let the angle of elevation from each corner be $	heta$.Then,$	an(	heta) = \frac{h}{d_1} = \frac{h}{d_2} = \frac{h}{d_3}$.This implies $d_1 = d_2 = d_3$.The point that is equidistant from all the vertices of a triangle is the circumcentre of the triangle.Therefore,the foot of the tower is at the circumcentre.

$A$ tower stands vertically inside an acute-angled triangular park $\Delta PQR$. If the angle of elevation of the top of the tower from each corner of the park is the same,then in $\Delta PQR$,the foot of the tower is at the

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