From the top of a tower, two boats on the same side of the tower are observed at angles of depression of 30° and 60°, respectively. If the distance between the boats is \(\frac{k \sqrt{3}}{3}\) m, then what is the height (in m) of the tower?

Given:

Angles of depression: 30° and 60°

Distance between boats: \(\frac{k \sqrt{3}}{3}\) m

Formula used:

tan θ = Opposite side / Adjacent side

Calculation:

Let the height of the tower be 'h' meters.

And the distance from the base of the tower to the nearer boat be 'x' meters.

Then, the distance from the base of the tower to the farther boat is x + (k√3)/3 meters.

From the 60° angle of depression:

tan 60° = h / x

⇒ √3 = h / x

⇒ x = h / √3

From the 30° angle of depression:

tan 30° = h / (x + k√3/3)

⇒ 1/√3 = h / (x + k√3/3)

⇒ x + k√3/3 = h√3

Substitute x = h/√3 into the equation:

⇒ h/√3 + k√3/3 = h√3

⇒ h/√3 - h√3 = -k√3/3

⇒ h(1/√3 - √3) = -k√3/3

⇒ h(1 - 3) / √3 = -k√3/3

⇒ h = (-k√3/3) × (-√3/2)

⇒ h = k(3/6) = k/2

∴ The height of the tower is k/2 meters.