Question

A statue 8 meters high standing on the top of a tower 64 meters high, on the bank of a river, subtends at a point A on the opposite bank, directly facing the tower, the same angle as subtended at the same point A by man of height 2 meters standing at the base of the tower, Prove that the breadth of the river is $16\sqrt{6}$ meters.

Answer 1

$tan\theta =\frac{2}{x}tan[\theta +\propto ]=\frac{72}{x}tan\propto =\frac{64}{x}tan[\theta +\propto ]=\frac{tan\theta +tan\propto }{1-tan\theta tan\propto }=\frac{y_n+\frac{64}{x}}{1-\frac{2}{x^2}.\frac{69}{{\pi }^2}}=\frac{66x}{x^2-128}$
$tan(\theta +\alpha )=\frac{65x}{x^2-128}=\frac{72}{x}66x62=72x^2-128\times 72128\times 72=6x^2{x}^2=\frac{128\times {72}^{12}}{6}=128\times 12$
$x=\sqrt{128\times 12}=\sqrt{4\times 4\times 4\times 2\times 4\times 3},=16\sqrt{6}proved.$
$\therefore$ Breadth of river =$16\sqrt{6}meters.$