Question

Question 15
A ladder against a vertical wall at an inclination $\alpha$ to the horizontal. Its foot is pulled away from the wall through a distance p, so that is upper end slides a distance q down the wall and then the ladder makes an angle $\beta$ to the horizontal. Show that $\frac{p}{q}=\frac{cos\beta -cos\alpha }{sin\alpha -sin\beta }$

Answer 1

Let OQ = x and OA = y

Given that, BQ = q, SA = p and AB = SQ = length of ladder

Also, $\angle BAQ=\alpha$ and $\angle QSO=\beta$

Now in $\Delta BAO$,

$cos\alpha =\frac{OA}{AB}$

$\Rightarrow cos\alpha =\frac{y}{AB}$

$\Rightarrow y=ABcos\alpha =OA\dots \dots \left(i\right)$

And $sin\alpha =\frac{OB}{AB}$

$\Rightarrow OB=BAsin\alpha$

Now, in $\Delta QSO$,

$cos\beta =\frac{OS}{SQ}$

$\Rightarrow OS=SQcos\beta =ABcos\beta [\because AB=SQ]\dots \dots \left(iii\right)$

And $sin\beta =\frac{OQ}{SQ}$

$\Rightarrow OQ=SQsin\beta =ABsin\beta$ $[\because AB=SQ]\dots \dots \left(iv\right)$

Now, SA = OS - AO

$P=ABcos\beta -ABcos\alpha$

$\Rightarrow P=AB(cos\beta -cos\alpha )\dots \dots \left(v\right)$

And BQ = BO - QO

$\Rightarrow q=BAsin\alpha -ABsin\beta$

$\Rightarrow q=AB(sin\alpha -sin\beta )\dots \dots \left(vi\right)$

Eq. (v) divided by Eq. (vi), we get

$\frac{p}{q}=\frac{AB(cos\beta -cos\alpha )}{AB(sin\alpha -sin\beta )}=\frac{cos\beta -cos\alpha }{sin\alpha -sin\beta }$

$\Rightarrow \frac{p}{q}=\frac{cos\beta -cos\alpha }{sin\alpha -sin\beta }$