Question

A ladder rests against a wall at an angle $\alpha$ to the horizontal. Its foot is pulled away from the wall through a distance a so that it slides a distanced down the wall making an angle $\beta$ with the horizontal, then

• A. $a=btan\frac{\alpha +\beta }{2}$
• B. $a=bcot\frac{\alpha +\beta }{2}$
• C. $atan\frac{\alpha -\beta }{2}$
• D. None

Answer 1

The correct option is A $a=btan\frac{\alpha +\beta }{2}$

In $\Delta OQB,cos\beta =\frac{OB}{BQ}$

⇒ $OB=\ell cos\beta ...........................\left(1\right)$

Similarly in $\Delta OPA,cos\alpha =\frac{OA}{PA}$

$\Rightarrow OA=\ell cos\alpha .........................\left(2\right)$

$Nowa=OB–OA=\ell (cos\beta –cos\alpha )..................\left(3\right)$

Also from $\Delta OAP,OP=\ell sin\alpha$

And in $△OQB;OQ=\ell sin\beta$

$\therefore b=OP–OQ=\ell (sin\alpha –sin\beta )......................\left(4\right)$

Dividing eq. $\left(3\right)$ by $\left(4\right)$ we get

$\frac{a}{b}=\frac{cos\beta –cos\alpha }{sin\alpha –sin\beta }$
$=\frac{2sin\frac{\alpha +\beta }{2}.sin\frac{\alpha -\beta }{2}}{2cos\frac{\alpha +\beta }{2}.sin\frac{\alpha -\beta }{2}}$

⇒ $\frac{a}{b}=tan\frac{(\alpha +\beta )}{2}$

Thus , $a=b$ $tan\frac{(\alpha +\beta )}{2}$ is proved