is a triangular park with m. A TV tower stands at the mid-point of . If the angles of elevation of the top of the tower at , and are respectively , and , then the height of the tower (in m) is
- A.100
- B.50
- C.
- D.
Show answer and solution
Correct option: A
Solution
Let M be the midpoint of QR where the tower stands, height h.
From P: tan45° = h/PM → PM = h.
From Q: tan30° = h/QM → QM = h√3.
Since PQ = 200 m and PM ⊥ QR (because PQR is isosceles with PQ=PR, so the median from P to QR is also perpendicular), by Pythagoras:
PQ² = PM² + QM²
200² = h² + 3h²= 4h²
h² = 10000
h = 100 m.
From P: tan45° = h/PM → PM = h.
From Q: tan30° = h/QM → QM = h√3.
Since PQ = 200 m and PM ⊥ QR (because PQR is isosceles with PQ=PR, so the median from P to QR is also perpendicular), by Pythagoras:
PQ² = PM² + QM²
200² = h² + 3h²= 4h²
h² = 10000
h = 100 m.
[JEE Main 2018]