From the top of a tower, the angles of depression of two objects A and B (situated on the ground on the same side of the tower) are observed to be 30° and 60°, respectively. If the distance between the objects is 200√3 m, then the height of the tower is?

Let height of tower = $h$ meters.

Since the angle of depression to object B is 60° and to object A is 30°, and 60° > 30°, object B is closer to the tower than object A.

The angle of depression from the tower equals the angle of elevation from the ground (alternate angles).

From object B, angle of elevation = 60°

From object A, angle of elevation = 30°

---

Let distance from tower base to object B = $x$

Let distance from tower base to object A = $y$

For object B:

$\tan(60°) = \dfrac{h}{x}$

$\sqrt{3} = \dfrac{h}{x}$

$x = \dfrac{h}{\sqrt{3}}$

---

For object A:

$\tan(30°) = \dfrac{h}{y}$

$\dfrac{1}{\sqrt{3}} = \dfrac{h}{y}$

$y = h\sqrt{3}$

---

Since B is closer and A is farther, the distance between them:

$y - x = 200\sqrt{3}$

$h\sqrt{3} - \dfrac{h}{\sqrt{3}} = 200\sqrt{3}$

$h\left(\sqrt{3} - \dfrac{1}{\sqrt{3}}\right) = 200\sqrt{3}$

$h\left(\dfrac{3}{\sqrt{3}} - \dfrac{1}{\sqrt{3}}\right) = 200\sqrt{3}$

$h\left(\dfrac{2}{\sqrt{3}}\right) = 200\sqrt{3}$

---

$h = 200\sqrt{3} \times \dfrac{\sqrt{3}}{2}$

$h = \dfrac{200 \times 3}{2}$

$h = 300$ m

Therefore, the height of the tower is 300 m.