Question

A drone is flying at a height of \(h\) metres. At an instant it observes the angle of elevation of the top of an industrial turbine as \(60^\circ\) and the angle of depression of the foot of the turbine as \(30^\circ\). If the height of the turbine is \(200\, \text{metres}\), find the value of \(h\) and the distance of the drone from the turbine.
(Use \(\sqrt{3} = 1.73\))

Hint:

Use tangent for right triangles in elevation/depression problems and apply Pythagoras to find slant distance.

Answer 1

Let the drone be flying at height \(h\) m.
Let the horizontal distance between drone and turbine be \(x\) m. From the diagram:
- Height of turbine = 200 m
- Angle of elevation of top of turbine = \(60^\circ\)
- Angle of depression of foot of turbine = \(30^\circ\)
So, difference in height between drone and top of turbine: \[ \text{Top part: } \tan 60^\circ = \dfrac{200 - h}{x} \quad \Rightarrow \sqrt{3} = \dfrac{200 - h}{x} \quad \text{(1)} \] \[ \text{Bottom part: } \tan 30^\circ = \dfrac{h}{x} \quad \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{h}{x} \quad \text{(2)} \] From (2): \(x = h \sqrt{3}\) Substitute into (1): \[ \sqrt{3} = \dfrac{200 - h}{h \sqrt{3}} \Rightarrow \sqrt{3} \cdot h \sqrt{3} = 200 - h \Rightarrow 3h = 200 - h \Rightarrow 4h = 200 \Rightarrow h = 50 \] Wait, double-check: From (2): \(x = h \sqrt{3}\)
Sub into (1): \[ \sqrt{3} = \dfrac{200 - h}{h \sqrt{3}} \Rightarrow \sqrt{3} \cdot h \sqrt{3} = 200 - h \Rightarrow 3h = 200 - h \Rightarrow 4h = 200 \Rightarrow h = 50 \Rightarrow x = h \sqrt{3} = 50 \cdot 1.73 = 86.5 \] Now find the slant distance from drone to top of turbine (hypotenuse of triangle): \[ \text{Using Pythagoras: } d^2 = x^2 + (200 - h)^2 = (86.5)^2 + (150)^2 = 7482.25 + 22500 = 29982.25 \Rightarrow d = \sqrt{29982.25} \approx 173\, \text{m} \]