Question

A ladder rests against a wall at an angle θ with the horizontal. When its foot is pulled away from the wall through a distance of y units, it slides a distance of x units down the wall and makes an angle $\alpha$ with the horizontal, then the value of $\frac{x}{y}$ is equal to

• A.
  $\tan \alpha -tan\theta$
• B.
  $\frac{\cos \theta -\cos \alpha }{\sin \theta -\sin \alpha }$
  
• C.
  $\frac{\sin \theta -\sin \alpha }{\cos \alpha -\cos \theta }$
  
• D.
  $\frac{\tan \alpha -\tan \theta }{\tan \alpha +\tan \theta }$
  

Answer 1

The correct option is C

$\frac{\sin \theta -\sin \alpha }{\cos \alpha -\cos \theta }$

Let the ladder AB be placed against the wall OA.

Here, AB = CD
$In\Delta AOB,$
$\sin \theta =\frac{AO}{AB}$
$\Rightarrow AO=AB\sin \theta$ ….(i)
And, $\cos \theta =\frac{OB}{AB}$
$\Rightarrow OB=AB\cos \theta$ ….(ii)
$In\Delta COD,$
$\sin \alpha =\frac{OC}{CD}$
$\Rightarrow OC=CD\sin \alpha =AB\sin \alpha$ $(\because AB=CD)$ ….(iii)
And, $\cos \alpha =\frac{OD}{CD}$
$\Rightarrow OD=CD\cos \alpha =AB\cos \alpha (\because AB=CD)$ ….(iv)
$\therefore \frac{x}{y}=\frac{AO-OC}{OD-OB}$
$\Rightarrow \frac{x}{y}=\frac{AB\sin \theta -AB\sin \alpha }{AB\cos \alpha -AB\cos \theta }$ [From (i),(ii),(iii) and (iv)]
$\Rightarrow \frac{x}{y}=\frac{\sin \theta -\sin \alpha }{\cos \alpha -\cos \theta }$
Hence, the correct answer is option (c).