Question

A plane mirror 50 cm long, is hung on a vertical wall of a room,with its lower edge 50 cm above the ground.A man stands infront of the mirror at a distance 2 m away from the mirror. If his eyes are at a height 1.8 m above the ground,then the length (distance between the extreme points of the visible region perpendicular to the mirror) of the floor visible to him due to reflection from the mirror is $\frac{x}{26}$ m. Find the value of x.

Answer 1

$\tan \theta =\frac{1}{y}$ and $\tan B=\frac{1/2}{x}$

also $\tan \theta =\frac{0.8}{2}=\frac{2}{5}$ and $\tan B=\frac{0.8+0.5}{2}=\frac{13}{20}$

$\therefore y=\frac{5}{2}$ and $x=\frac{1}{2\times \frac{13}{20}}=\frac{10}{13}$

$\therefore$ length of floor visible to him $=y-x=\frac{5}{2}-\frac{10}{13}=\frac{65-20}{26}=\frac{45}{26}$

$\therefore$ as per question $x=45$