is a triangular park with m. A vertical tower is situated at the mid-point of . If the angles of elevation of the top of the tower at and are and respectively, then the height of the tower (in m) is
- A.25
- B.20
- C.
- D.
Show answer and solution
Correct option: B
Solution
Let M be the midpoint of BC, h the tower height. AM ⊥ BC (since AB=AC).
Let BM = a, so BC = 2a.
From A: tan(angle at A) = h/AM, where angle = cot⁻¹(3√2), so cot = 3√2, tan = 1/(3√2).
Thus AM = 3√2·h.
From B: angle = cosec⁻¹(2√2), so cosec = 2√2, sin = 1/(2√2), and the hypotenuse BM² + h² gives the slant distance from B to top. Actually BM is the horizontal distance from B to tower base M.
cosec(angle at B) = 2√2 means sin = 1/(2√2), so the slant distance / h = 2√2 (since sin θ = h/slant), giving slant = 2√2·h.
Thus BM² + h² = (2√2·h)² = 8h², so BM² = 7h².
Now in triangle ABM: AB² = AM² + BM² (right angle at M):
100² = (3√2·h)² + 7h² = 18h² + 7h² = 25h²
h² = 10000/25 = 400, so h = 20 m.
Let BM = a, so BC = 2a.
From A: tan(angle at A) = h/AM, where angle = cot⁻¹(3√2), so cot = 3√2, tan = 1/(3√2).
Thus AM = 3√2·h.
From B: angle = cosec⁻¹(2√2), so cosec = 2√2, sin = 1/(2√2), and the hypotenuse BM² + h² gives the slant distance from B to top. Actually BM is the horizontal distance from B to tower base M.
cosec(angle at B) = 2√2 means sin = 1/(2√2), so the slant distance / h = 2√2 (since sin θ = h/slant), giving slant = 2√2·h.
Thus BM² + h² = (2√2·h)² = 8h², so BM² = 7h².
Now in triangle ABM: AB² = AM² + BM² (right angle at M):
100² = (3√2·h)² + 7h² = 18h² + 7h² = 25h²
h² = 10000/25 = 400, so h = 20 m.
[JEE Main 2019 · 10 Apr · Shift I]