Question : Two pillars A and B of the same height are on opposite sides of a road which is 40 m wide. The angles of elevation of the tops of the pillars A and B are $30^{\circ}$ and $45^{\circ}$, respectively, at a point on the road between the pillars. What is the distance (in m ) of the point from the foot of pillar A?
Option 1: $40(\sqrt{3}-1)$
Option 2: $20(2-\sqrt{3})$
Option 3: $20(3-\sqrt{3})$
Option 4: $39 \sqrt{3}$
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Correct Answer: $20(3-\sqrt{3})$
Solution : In $\triangle ROS$, $\tan45^\circ = \frac{h}{OS}$ ⇒ $1= \frac{h}{OS}$ $OS = h$ Thus, $QO = (40 - h)$ Now, In $\triangle POQ$, $\tan30^\circ = \frac{h}{40 - h}$ ⇒ $\frac{1}{\sqrt 3} = \frac{h}{(40 - h)}$ ⇒ $40 - h = \sqrt 3 h$ ⇒ $h (\sqrt 3 + 1) = 40$ ⇒ $h = \frac{40}{(\sqrt 3 + 1)}$ $QO = (40 - h)$ $QO=40 - \frac{40}{(\sqrt 3 + 1)}$ $QO = \frac{(40\sqrt 3)}{(\sqrt 3 + 1)}$ After rationalisation, we get: $QO = \frac{(40\sqrt 3)}{(\sqrt 3 + 1)}\times \frac{(\sqrt 3 - 1)}{(\sqrt 3 - 1)}$ $QO = 20(3 - \sqrt 3)$ Hence, the correct answer is $20(3 - \sqrt 3)$.
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