Question

An aeroplane when flying at a height of 3000 metres from the ground passes vertically above another aeroplane at an instant when the angles of elevation of the two planes from the same point on the ground are ${60}^{\circ }$ and ${45}^{\circ }$ respectively. Find the vertical distance between the aeroplanes at that instant. [Take $\sqrt{3}=1.73$] [3 MARKS]

Answer 1

Let O be the point of observation. Let A and B be the positions of the two planes at the given instant when A is vertically above B.



Let AB when produced meet the ground at C.

Then, $\angle COA={60}^{\circ },\angle COB={45}^{\circ },\angle OCB=\angle OCA={90}^{\circ }$ and AC = 3000 m.

Let AB = x metres. Then, BC = (3000 - x) m

From right $\Delta OCB$, we have

$\frac{OC}{BC}=cot{45}^{\circ }=1\Rightarrow \frac{OC}{(3000-x)m}=1$

$\Rightarrow OC=(3000-x)m$ ............ (i)

From right $\Delta OCA$, we have

$\frac{OC}{AC}=cot{60}^{\circ }=\frac{1}{\sqrt{3}}\Rightarrow \frac{OC}{3000m}=\frac{1}{\sqrt{3}}$

$\Rightarrow OC=(\frac{3000}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}})m=1000\sqrt{3}m$ ............ (ii)

From (i) and (ii), we get

$(3000-x)=1000\sqrt{3}$

$\Rightarrow x=(3000-1000\sqrt{3})=(3000-1000\times 1.73)$

$=(3000-1730)=1270$

Hence, the required distance between the two aeroplanes is 1270 metres.

[$\frac{1}{2}$ MARK]

In $\Delta ADC$

$tan60=\frac{3000}{x}$

$\sqrt{3}=\frac{3000}{x}$

$x=\frac{3000}{\sqrt{3}}\dots \dots \left(1\right)$ [1 MARK]

In $\Delta BDC$

$tan45=\frac{h}{x}$

$1=\frac{h}{x}$

$h=x=\frac{3000}{\sqrt{3}}$ [1 MARK]

Distance between the planes

$=3000-\frac{3000}{\sqrt{3}}$

$=3000\left(\frac{\sqrt{3}-1}{\sqrt{3}}\right)$

$=\text{/}{3000}^{1000}\frac{(3-\sqrt{3})}{\text{/}3}$

$=1000\left(1.268\right)$

= 1268 m [1 $\frac{1}{2}$ MARK]