Question

If the angle of elevation top of point $D$ of leaning tower at the point $A$ and $B$ is $\alpha$ and $\beta$. If slope of tower is $\theta$, then prove that
$\cot \theta =\frac{b\cot \alpha -\alpha \cot \beta }{b-a}$
Where $a$ and $b$ are the distances of $A$ and $B$ from tower $(b>a)$.

Answer 1

Let $CD$ is a leaning tower and angles of elevations of top of point $D$ at the place $A$ and $B$ are $\alpha$ and $\beta$.

So, $\angle DBM=\beta ,\angle DAM=\alpha$

Let $\angle DCM=\theta$ and $DM=h$ m, $CM=x$ m and $AC=a,BC=b$.

From right angled $\Delta DMC$,

$\tan \theta =\frac{DM}{CM}$

$\tan \theta =\frac{h}{x}$

$x=\frac{h}{\tan \theta }=h\cot \theta$ ….(i)

From right angled $\Delta DMA$,

$\tan \alpha =\frac{DM}{AM}$

$\tan \alpha =\frac{h}{a+x}$

$a+x==h\cot \alpha$

$a=h\cot \alpha –x$

$=h\cot \alpha –h\cot \theta$

$=h[\cot \alpha –\cot \theta ]$ ……(ii)

For right angled $\Delta DMB$,

$\tan \beta =\frac{DM}{BM}$

$\cot \beta =\frac{BM}{DM}$

$\cot \beta =\frac{b+x}{h}$

$b+x=h\cot \beta$

$b=h\cot \beta –x$

$b=h\cot \beta –h\cot \theta$

$=h[\cot \beta –\cot \theta ]$ ……(iii)

Divide equation (iii) in equation (ii),

$\frac{a}{b}=\frac{h(\cot \alpha -\cot \theta )}{h(\cot \beta -\cot \theta )}$

$a(\cot \beta –\cot \theta )=b(\cot \alpha –\cot \theta )$

$(b–a)\cot \theta =b\cot \alpha –a\cot \beta$

$\cot \theta =\frac{b\cot \alpha -\alpha \cot \beta }{b-a}$