Two pillars A and B of the same height are on opposite sides of a road which is 100 m wide. The angles of elevation of the top of the pillars A and B are 60° and 45°  respectively, from a point on the road between the pillars. What is the distance (in m) of the point from the foot of pillar A?

50(√3 - 1) m

Let AB = CD = (x) m

In ΔABQ:

⇒ tan 60° = \(\frac{AB}{BQ}\)

⇒ \(\frac{\sqrt {3}}{1}\) = \(\frac{AB}{BQ}\)

Here, AB = \(\sqrt {3}\) and BQ = 1 ................(i)

In ΔCDQ:

⇒ tan 45° = \(\frac{CD}{DQ}\)

⇒ \(\frac{1}{1}\) = \(\frac{CD}{DQ}\)

⇒ \(\frac{\sqrt {3}}{\sqrt {3}}\) = \(\frac{CD}{DQ}\)    [because CD = AB]

Here, CD = \(\sqrt {3}\) and DQ = \(\sqrt {3}\) .................(ii)

From (i) and (ii)

⇒ DB = DQ + QB = [\(\sqrt {3}\) + 1]R

⇒ [\(\sqrt {3}\) + 1]R = 100 (given)

⇒ 1R = \(\frac{100}{\sqrt {3}\;+\;1}\) = \(\frac{100}{2}\) (\(\sqrt {3}\) - 1) = 50 (\(\sqrt {3}\) - 1)

So,

Distance between point B and Q = 1R = 50 (\(\sqrt {3}\) - 1) m