Question

A man sees the top of a tower in a mirror which is at a distance of $87.6m$ from the tower. The mirror is on the ground, facing upward. The man is $0.4m$ away from the mirror, and the distance of his eye level from the ground is $1.5m$. How tall is the tower? (The foot of man, the mirror and the foot of the tower lie along a straight line)

Answer 1

Let $AB$ and $ED$ be the heights of the man and the tower respectively. Let $C$ be the point of incidence of the tower in the mirror.
In $△ABC$ and $△EDC$, we have
$\angle ABC=\angle EDC={90}^{\circ }$
$\angle BCA=\angle DCE$
(angular elevation is same at the same instant. i.e., the angle of incidence and the angle of reflection are same.)
$\therefore △ABC\sim △EDC$ ($AA$ similarity criterion)
Thus, $\frac{ED}{AB}=\frac{DC}{BC}$ (corresponding sides are proportional)
$ED=\frac{DC}{BC}\times AB=\frac{87.6}{0.4}\times 1.5=328.5$
Hence, the height of the tower is $328.5m$.