Raj stands in a corner of his square farm. Angle of elevation of a scarecrow placed in diagonally opposite corner is 60°. He starts walking backwards in a straight line and after 80ft he realizes that angle of elevation of the scarecrow now is 30°. What is area of the field?

Solution(By Examveda Team)

$$\eqalign{ & \tan {60^ \circ } = \sqrt 3 = \frac{{PQ}}{{QR}} \cr & \therefore PQ = \sqrt 3 \,QR \cr & \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{PQ}}{{SQ}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{PQ}}{{80 + QR}} \cr & \therefore 80 + QR = \sqrt 3 \,PQ \cr & \therefore 80 + QR = 3QR \cr & \therefore QR = 40\,{\text{ft}}{\text{.}} \cr} $$
If we read carefully, we see that Raj (Point R) and the scarecrow (Point Q) are in diagonally opposite corners.
So QR is a diagonal of the square farm.
Diagonal of square = side x$$\sqrt 2 $$
$$\eqalign{ & \therefore 40 = {\text{side}}\,{\text{x}}\sqrt 2 \cr & \therefore {\text{side}} = \frac{{40}}{{\sqrt 2 }} \cr & \therefore {\text{Area}} = {\left( {{\text{side}}} \right)^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left( {\frac{{40}}{{\sqrt 2 }}} \right)^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {\frac{{1600}}{2}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 800\,{\text{sq}}{\text{.}}\,{\text{ft}}{\text{.}} \cr} $$