Question

The angle of elevation of a jet plane from a point A on the ground is ${60}^{\circ }$. After a flight of 30 seconds, the angle of elevation changes to ${30}^{\circ }$. If the jet plane is flying at a constant height of $3600\sqrt{3}metres$, find the speed of the jet plane. [4 MARKS]

Answer 1

Let A be the point of observation and let AX be a horizontal line through A and $QC\perp AX$. Let P and Q be the two positions of the plane. Let $PB\perp AX$.



Then, $PB=QC=3600\sqrt{3}metres$, $\angle BAP={60}^{\circ }$ and $\angle BAQ={30}^{\circ }$

From right $\Delta ABP$, we have

$\frac{AB}{BP}=cot{60}^{\circ }=\frac{1}{\sqrt{3}}\Rightarrow \frac{AB}{3600\sqrt{3}}=\frac{1}{\sqrt{3}}$

$\Rightarrow AB=3600\sqrt{3}\times \frac{1}{\sqrt{3}}=3600m$ ...... (i)

Let $BC=PQ=xmetres$

Then, $AC=AB+BC=(x+3600)m$ [using (i)]

From right $\Delta ACQ$, we have

$\frac{AC}{CQ}=cot{30}^{\circ }=\sqrt{3}\Rightarrow \frac{x+3600}{3600\sqrt{3}}=\sqrt{3}$

$\Rightarrow x+3600=(3600\times 3)=10800$

$\Rightarrow x=10800-3600=7200$

Thus, PQ = 7200 metres

Now, 7200 m is covered in 30 seconds.

$\therefore$ speed of the jet plane $=(\frac{7200}{30}\times \frac{60\times 60}{1000})km/hr$ (Here, We convert metre per second to kilometre per hour)

$=864km/hr$.

Hence, the speed of the jet plane is 864 km/hr.

Alternative Method,

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

In $\Delta ABC$

$tan60=\frac{3600\sqrt{3}}{x}$

$\sqrt{3}=\frac{3660\sqrt{3}}{x}$

$x=3600\dots \dots \left(1\right)$ [1 MARK]

In $\Delta ABE$

$tan30=\frac{3600\sqrt{3}}{y}$

$\frac{1}{\sqrt{3}}=\frac{3600\sqrt{3}}{y}$

$y=3600\left(3\right)$ [1 MARK]

To travel BD or (y-x) it takes 30 sec [$\frac{1}{2}$ MARK]

i.e. speed $=\frac{\left[3600\right(3)-3600]m}{30}$

$=\frac{\text{/}{3600}^{120}\left(2\right)}{\text{/}30}$

$=240m/s$

$=\frac{\text{/}240}{\text{/}1000}\times \text{/}60\times \text{/}60$
(Here, We convert metre per second to kilometre per hour)
$=864km/h$