If the angles of elevation of the top of a tower from three collinear points $$A$$, $$B$$ and $$C$$ on a line leading to the foot of the tower are $$30^\circ$$, $$45^\circ$$ and $$60^\circ$$ respectively, then the ratio $$AB : BC$$, is

Let us denote the foot of the tower by the point $$O$$ and the top of the tower by the point $$T$$. The height of the tower is unknown, so we call it $$h = OT$$. The three points $$A,\,B,\,C$$ lie on the straight line that passes through $$O$$, with $$A$$ being the farthest from the tower and $$C$$ the closest, because the angle of elevation increases as we approach the tower.

We measure the horizontal distances from each point to the foot $$O$$ and name them $$OA = x_A,\qquad OB = x_B,\qquad OC = x_C.$$

From each of the three points we are given an angle of elevation to the top $$T$$: $$\angle ATO = 30^\circ,\qquad \angle BTO = 45^\circ,\qquad \angle CTO = 60^\circ.$$

Whenever we have a right-angled triangle with an angle $$\theta$$, the tangent formula states $$\tan\theta = \frac{\text{opposite side}}{\text{adjacent side}}.$$ Here the “opposite side’’ is always the height $$h$$ and the “adjacent side’’ is the corresponding horizontal distance. We therefore write for each point:

From point $$A$$: $$\tan 30^\circ = \frac{h}{x_A}.$$

From point $$B$$: $$\tan 45^\circ = \frac{h}{x_B}.$$

From point $$C$$: $$\tan 60^\circ = \frac{h}{x_C}.$$

Now we substitute the well-known numerical values $$\tan 30^\circ = \frac{1}{\sqrt3},\qquad \tan 45^\circ = 1,\qquad \tan 60^\circ = \sqrt3.$$

Hence we have the three equations

$$\frac{1}{\sqrt3} = \frac{h}{x_A} \;\;\Longrightarrow\;\; x_A = h\sqrt3,$$

$$1 = \frac{h}{x_B} \;\;\Longrightarrow\;\; x_B = h,$$

$$\sqrt3 = \frac{h}{x_C} \;\;\Longrightarrow\;\; x_C = \frac{h}{\sqrt3}.$$

Our required distances along the ground are the differences:

Distance $$AB$$: $$AB = OA - OB = x_A - x_B = h\sqrt3 - h = h(\sqrt3 - 1).$$

Distance $$BC$$: $$BC = OB - OC = x_B - x_C = h - \frac{h}{\sqrt3} = h\!\left(1 - \frac{1}{\sqrt3}\right) = h\!\left(\frac{\sqrt3 - 1}{\sqrt3}\right).$$

We now form the desired ratio $$AB:BC$$. First write both distances explicitly:

$$AB = h(\sqrt3 - 1),\qquad BC = h\!\left(\frac{\sqrt3 - 1}{\sqrt3}\right).$$

Because both terms contain the common factor $$h(\sqrt3 - 1)$$, we can cancel it:

$$\frac{AB}{BC} = \frac{h(\sqrt3 - 1)}{\,h(\sqrt3 - 1)/\sqrt3} = \frac{h(\sqrt3 - 1)}{h(\sqrt3 - 1)}\;\times\;\sqrt3 = \sqrt3.$$

Therefore the simplified ratio is

$$AB : BC = \sqrt3 : 1.$$

Hence, the correct answer is Option B.