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?
Answer & explanation
Correct answer: option 1
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