Question

A ladder of length $k$ rests against a wall at an angle $\alpha$ to the horizontal. If foot is pulled away through a distance a, so that is slides a distance b down the wall, finally making an angle $\beta$ with the horizontal. Then

• A. $\tan \left(\frac{\alpha +\beta }{2}\right)=\frac{a}{b}$
• B. $\tan \left(\frac{\alpha -\beta }{2}\right)=\frac{a}{b}$
• C. $cosec\left(\frac{\alpha -\beta }{2}\right)=\frac{2k}{\sqrt{a^2+b^2}}$
• D. $\sin (\alpha +\beta )=\frac{ab}{k^2}$

Answer 1

Answer: (B), (C)

The correct options are
B $\tan \left(\frac{\alpha -\beta }{2}\right)=\frac{a}{b}$
C $cosec\left(\frac{\alpha -\beta }{2}\right)=\frac{2k}{\sqrt{a^2+b^2}}$

$AB=A^{\prime }{B}^{\prime }=k$ (length of ladder)

$A^{\prime }A=a,B{B}^{\prime }=b$

$a=O{A}^{\prime }-OA,b=OB-O{B}^{\prime }$.......(I)

In ${\Delta }^{le}OAB$

$\cos \alpha =\frac{OA}{AB}=\frac{OA}{K}\Rightarrow OA-k\cos \alpha$

$\sin \alpha =\frac{OB}{AB}=\frac{OB}{K}\Rightarrow OB-k\sin \alpha$

In ${\Delta }^{le}O{A}^{\prime }{B}^{\prime }$

$\cos \beta =\frac{O{A}^{\prime }}{A^{\prime }{B}^{\prime }}=\frac{O{A}^{\prime }}{K}\Rightarrow O{A}^{\prime }-k\cos \beta$

$\sin \beta =\frac{O{B}^{\prime }}{A^{\prime }{B}^{\prime }}=\frac{O{B}^{\prime }}{K}\Rightarrow O{B}^{\prime }-k\sin \beta$

Sub above equation in(I)

$\alpha =k\cos \beta -k\cos \alpha$ $b=k\sin \alpha -k\sin \beta$

$\alpha =k(\cos \beta -\cos \alpha )$ $b=k(\sin \alpha -k\sin \beta )$

$\frac{a}{b}=\frac{k(\cos \beta -\cos \alpha )}{k(\sin \alpha -\sin \beta )}$

$\frac{a}{b}=\frac{\cos \beta -\cos \alpha }{\sin \alpha -\sin \beta }$

$=\frac{2.\sin \left(\frac{\beta +\alpha }{2}\right)\sin \left(\frac{\alpha +\beta }{2}\right)}{2.\sin \left(\frac{\alpha -\beta }{2}\right).\cos \left(\frac{\alpha -\beta }{2}\right)}$

$\frac{a}{b}=\tan \left(\frac{\alpha -\beta }{2}\right)$

$a^2=k^2(\cos \beta -\cos \alpha {)}^2$

$b^2=k^2(\sin \alpha -\sin \beta {)}^2$

$a^2+b^2=k^2[{\cos }^2\beta +{\cos }^2\alpha -2\cos \alpha \cos \beta +{\sin }^2\alpha +{\sin }^2\beta -2\sin \alpha \sin \beta ]$

$=k^2[{\sin }^2\alpha +{\cos }^2\alpha +{\sin }^2\beta +\cos d^2\beta -2(\cos \alpha \cos \beta +\sin \alpha +\sin \beta \left)\right]$

$=k^2[1+1-2(\cos (\alpha -\beta )\left)\right]$

$k^2[2-2\cos (\alpha -\beta \left)\right]$

$=2k^2[1-\cos (\alpha -\beta \left)\right]$

$=2k^2[1-(1-(1-2{\sin }^2(\frac{\alpha -\beta }{2}\left)\right)\left)\right]$

$=\frac{a^2+b^2}{2k^2}=1-1+2{\sin }^2\left(\frac{\alpha -\beta }{2}\right)$

$=\frac{a^2+b^2}{2k^2}=2{\sin }^2\left(\frac{\alpha -\beta }{2}\right)$

$\frac{a^2+b^2}{4k^2}={\sin }^2\left(\frac{\alpha -\beta }{2}\right)$

$cose{c}^2\left(\frac{\alpha -\beta }{2}\right)=\frac{1}{{\sin }^2\left(\frac{\alpha -\beta }{2}\right)}=\frac{4k^2}{a^2+b^2}$

$cosec\left(\frac{\alpha -\beta }{2}\right)=\sqrt{\frac{4k^2}{a^2+b^2}}$

$cosec\left(\frac{\alpha -\beta }{2}\right)=\frac{2k}{\sqrt{a^2+b^2}}$

$\therefore cosec\left(\frac{\alpha -\beta }{2}\right)=\frac{2k}{\sqrt{a^2+b^2}}$