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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