JEE PYQ: Heights And Distances Question 12
Question 12 - 2019 (10 Apr Shift 1)
ABC is a triangular park with $AB = AC = 100$ metres. A vertical tower is situated at the mid-point of BC. If the angles of elevation of the top of the tower at A and B are $\cot^{-1}(3\sqrt{2})$ and $\csc^{-1}(2\sqrt{2})$ respectively, then the height of the tower (in metres) is:
(1) $\frac{100}{3\sqrt{3}}$ (2) $10\sqrt{5}$ (3) $20$ (4) $25$
Show Answer
Answer: (3) $20$
Solution
$18h^2 + 7h^2 = 10000 \Rightarrow h = 20$.
BETA
