An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour.
Question.

An aeroplane leaves an airport and flies due north at a speed of $1000 \mathrm{~km}$ per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of $1200 \mathrm{~km}$ per hour. How far apart will be the two planes after $1 \frac{1}{2}$ hours?


Solution:

The first plane travels distance $\mathrm{BC}$ in the direction of north in $1 \frac{1}{2}$ hours at a speed of 1000 $\mathrm{km} / \mathrm{hr}$.

$\therefore \quad \mathrm{BC}=1000 \times \frac{3}{2} \mathrm{~km}=1500 \mathrm{~km} .$

The first plane travels distance BC

The second plane travels distance $\mathrm{BA}$ in the direction of west in $1 \frac{1}{2}$ hours at a speed of 1200 $\mathrm{km} / \mathrm{hr}$.

$\therefore \quad \mathrm{BA}=1200 \times \frac{3}{2} \mathrm{~km}=1800 \mathrm{~km} .$

From right angled $\triangle \mathrm{ABC}$,

$\mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2}$

$=(1800)^{2}+(1500)^{2}$

$=3240000+2250000=5490000$

$\Rightarrow \mathrm{AC}=\sqrt{5490000} \mathrm{~m} \Rightarrow \mathrm{AC}=300 \sqrt{61} \mathrm{~m}$
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