Question

A statue 20 m. High standing at the top of a column 150 m, high on the bank of a river, subtends at a point on the opposite bank directly facing the column, the same angle as subtended at the same point by a man 2 m high standing at the base of the column. Find the breadth of the river.

Answer 1

AD = 2, AB = 150, BC = 20
Let PA = x
$2=xtan\beta ;\Delta PDA$
$150=xtan\alpha ;\Delta PBA$
$170=xtan(\alpha +\beta )$
$tan(\alpha +\beta )=\frac{152x}{x^2-300}$
or $152x^2=170x^2-51000$
or $18x^2=51000$
or $x^2=\frac{25500}{9}$
$\therefore x=\frac{10}{3}\sqrt{255}$
$=\frac{160}{3}$
= 53 metres approx.