Answer:
Let BC be the height at which the aeroplane flying. Then, \[BC=1500\sqrt{3}m\] In 15 seconds, the aeroplane moves from C to E and makes angle of elevation \[30{}^\circ \]. Let \[AB=x\text{ }m,\text{ }BD=y\text{ }m\] So, \[AD=(x+y)\text{ }m\] In \[\Delta \,ABC\], \[\tan \,60{}^\circ =\frac{BC}{AB}\] \[\sqrt{3}=\frac{1500\sqrt{3}}{x}\] \[[\because \,tan\,60{}^\circ =\sqrt{3}]\] \[x=1500\,\,m\] ?(i) In \[\Delta \text{ }EAD\] \[tan\text{ }30{}^\circ =\frac{ED}{AD}\] \[\left[ \because \,\tan \,30{}^\circ =\frac{1}{\sqrt{3}} \right]\] \[\frac{1}{\sqrt{3}}=\frac{1500\sqrt{3}}{x+y}\] \[x+y=1500\times 3\] \[y=4500-1500=3000\,m\] [Using equation (i)] Speed of aeroplane\[=\frac{Distance}{Time}\] \[=\frac{3000}{15}\] \[=200\,m/s\,\] or \[720\,km/hr.\]
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