question_answer

The angle of elevation of an aeroplane from point A on the ground is \[60{}^\circ \]. After flight of 15 seconds, the angle of elevation change to \[30{}^\circ \]. If the aeroplane is flying at a constant height of \[1500\sqrt{3}\,m\], find the speed of the plane in km/hr.

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 \[720\,km/hr.\]