A person stands at a point $A$ due south of a tower and observes elevation $60^\circ$. He walks west to $B$, elevation becomes $45^\circ$. At point $C$ on $AB$ extended, elevation becomes $30^\circ$. Find $\dfrac{AB}{BC}$.

Let height of tower = $h$

From A (angle $60^\circ$):

$\tan 60^\circ = \frac{h}{OA} \Rightarrow OA = \frac{h}{\sqrt{3}}$

From B (angle $45^\circ$):

$\tan 45^\circ = \frac{h}{OB} \Rightarrow OB = h$

$AB = OB - OA = h - \frac{h}{\sqrt{3}}$

From C (angle $30^\circ$):

$\tan 30^\circ = \frac{h}{OC} \Rightarrow OC = \sqrt{3}h$

$BC = OC - OB = \sqrt{3}h - h$

Now ratio:

$\frac{AB}{BC} = \frac{h - \frac{h}{\sqrt{3}}}{\sqrt{3}h - h}$

$= \frac{h\left(1 - \frac{1}{\sqrt{3}}\right)}{h(\sqrt{3} - 1)}$

$= \frac{1 - \frac{1}{\sqrt{3}}}{\sqrt{3} - 1}$

Multiply numerator & denominator by $\sqrt{3}$:

$= \frac{\sqrt{3} - 1}{\sqrt{3}(\sqrt{3} - 1)}$

$= \frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{1} = 1$

Final Answer: $$\boxed{1}$$