Question

The top of a tower is observed from three points $A,B,C$ on a straight line leading to the tower. If the angles of elevation are $\theta ,2\theta ,3\theta$ from them prove that
$AB:BC::(\cot \theta -\cot 2\theta ):(\cot 2\theta -\cot 3\theta )$.
If $C$ is found to be at the foot of the tower prove that $B$ trisects $AC$.

Answer 1

From the figure of $Q.14\left(a\right)$, we get
$\frac{AB}{BC}=\frac{AD-BD}{BD-CD}=\frac{h(\cot \theta -\cot 2\theta )}{h(\cot 2\theta -\cot 3\theta )}$ ...$\left(1\right)$
or $\frac{AB}{BC}=\frac{\cot \theta -\cot 2\theta }{\cot 2\theta \cot 3\theta }$
Note: In case $C$ is the root, then $3\theta ={90}^o$
$\therefore \theta =30,$ and $\cot {30}^o=\sqrt{3}$,$\cot 90=0$
$\frac{AB}{BC}=\frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{\frac{1}{\sqrt{3}}-0}=\frac{2}{1}$
or $AB=2BC$.
Hence $B$ trisects $AC$.