A ladder rests against a vertical wall at an inclination α to the horizontal. Its foot is pulled away from the wall through a distance p so that its upper end slides a distance q down the wall and then the ladder makes an angle β to the horizontal. Show t

Name: A ladder rests against a vertical wall at an inclination α to the horizontal. Its foot is pulled away from the wall through a distance p so that its upper end slides a distance q down the wall and then the ladder makes an angle β to the horizontal. Show t
Uploaded: Mon, 2021-02-01 18:17
Description: A ladder rests against a vertical wall at an inclination α to the horizontal. Its foot is pulled away from the wall through a distance p so that its upper end slides a distance q down the wall and then the ladder makes an angle β to the horizontal. Show t

A ladder rests against a vertical wall at an inclination α to the horizontal. Its foot is pulled away from the wall through a distance p so that its upper end slides a distance q down the wall and then the ladder makes an angle β to the horizontal. Show that $\frac{p}{q}= \frac{\cos \beta -\cos \alpha }{\sin \alpha-\sin \beta }$

Answers (1)

Here a and b are the angles indicating when the ladder is at rest and when it is pulled away from the wall.
In $\bigtriangleup$ AOB
$\cos \alpha = \frac{OB}{AB}$              $cos \theta = \frac{Base}{Hypotenuse}$
$OB= AB\cos \alpha \cdots \left ( 1 \right )$
$\sin \alpha = \frac{AO}{AB}$                $\sin \theta = \frac{Perpendicular}{hypotenuse}$

$AO= AB\sin \alpha \cdots \left ( 2 \right )$
Similarly In $\bigtriangleup$ DOC
$\cos \beta = \frac{OC}{DC}$
$OC= DC\cos \beta \cdots \left ( 3 \right )$
$\sin \beta = \frac{OD}{DC}$
$OD= DC\sin \beta \cdots \left ( 4 \right )$
Now subtract equation (1) from (3) we get
OC – OB = DC cos $\beta$ – AB cos $\alpha$
Here    OC – OB = P
and DC = AB because length of ladder remains $\Rightarrow$ P = AB cos
$\beta$ – AB cos $\alpha$
    P = AB (cos $\beta$ – cos $\alpha$ )         …(5)
Subtract equation (4) from (2) we get
AO – OD = AB sina – DC sin $\beta$
Here AO – OD = q
and AB = DC because the length of the ladder remains the same
     $\Rightarrow$    q = AB sin $\alpha$ – AB sin $\beta$
q = AB (sin $\alpha$ – sin $\beta$ )             …(6)
By dividing equations (5) and (6) we get
$\frac{p}{q}= \frac{AB\left ( \cos \beta -\cos \alpha \right )}{AB\left ( \sin \alpha -\sin \beta \right )}$
$\frac{p}{q}= \frac{\cos \beta -\cos \alpha}{\sin \alpha -\sin \beta}$
Hence proved