JEE PYQ: Heights And Distances Question 13
Question 13 - 2019 (12 Apr Shift 2)
The angle of elevation of the top of a vertical tower standing on a horizontal plane is observed to be $45°$ from a point A on the plane. Let B be the point 30 m vertically above the point A. If the angle of the top of the tower from B be $30°$, then the distance (in m) of the foot of the tower from the point A is:
(1) $15(3+\sqrt{3})$ (2) $15(5-\sqrt{3})$ (3) $15(3-\sqrt{3})$ (4) $15(1+\sqrt{3})$
Show Answer
Answer: (1) $15(3+\sqrt{3})$
Solution
$d = \frac{30\sqrt{3}}{\sqrt{3}-1} = 15(3+\sqrt{3})$.
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