NIMCET Heights And Distances Previous Year Questions (PYQs) – Page 1 of 2

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: $$\boxed{1}$$