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The angle of elevation of an aeroplane from a point on the ground is ${{45}^{\circ }}$ . After a flight of 15 seconds, the elevation changes to ${{30}^{\circ }}$ . If the aeroplane flying at a height of 3000 meters, find the speed of the aeroplane.

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Last updated date: 28th Mar 2024
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MVSAT 2024
Answer
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Hint: First draw the figure. A rough sketch according to the information given in the question. Consider a triangle, and use $\text{tan }\!\!\theta\!\!\text{ =}\dfrac{\text{perpendicular}}{\text{base}}$ in that triangle to calculate the distance between point of observation and point where aeroplane reached with angle of elevation ${{45}^{\circ }}$ . Now, consider another triangle and use $\text{tan }\!\!\theta\!\!\text{ =}\dfrac{\text{perpendicular}}{\text{base}}$ in that triangle to calculate the distance between point of observation and point where aeroplane reached with angle of elevation ${{30}^{\circ }}$ . After that calculate the difference between these two calculated distances to get the distance travelled by aeroplane in 15 sec. Then divide this distance travelled by 15 sec to get the speed of the aeroplane in ${m}/{s}\;$ .

Complete step-by-step answer:

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Let the height of flying of the aeroplane be $PQ=BC$ and point A be the point of observation.

We have,

$\begin{align}

  & PQ=BC=3000m \\

 & \angle PAQ={{45}^{\circ }}\text{ and} \\

 & \angle BAC={{30}^{\circ }} \\

\end{align}$

Now, we consider $\Delta PAQ$ ,

So, in $\Delta PAQ$ ,

$\begin{align}

  & \tan \angle PAQ=\dfrac{PQ}{AQ} \\

 & \text{Since, }\angle PAQ={{45}^{\circ }} \\

 & \Rightarrow \tan {{45}^{\circ }}=\dfrac{PQ}{AQ} \\

\end{align}$

We have, $PQ=3000m$

$\begin{align}

  & \therefore \tan {{45}^{\circ }}=\dfrac{3000}{AQ} \\

 & \Rightarrow 1=\dfrac{3000}{AQ} \\

\end{align}$ (we know that, $\tan {{45}^{\circ }}=1$ )

$\therefore AQ=3000m$…………………………………(1)

Now, we consider $\Delta ABC$ ,

So, in $\Delta ABC$ ,

$\tan \angle BAC=\dfrac{BC}{AC}$ (As $\text{tan }\!\!\theta\!\!\text{ =}\dfrac{\text{perpendicular}}{\text{base}}$ ).

Since, we have here $\angle BAC={{30}^{\circ }}$ .

$\begin{align}

  & BC=PQ=3000m \\

 & \Rightarrow \tan {{30}^{\circ }}=\dfrac{BC}{AC} \\

 & \Rightarrow \tan {{30}^{\circ }}=\dfrac{3000}{AC} \\

 & \Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{3000}{AC} \\

\end{align}$ (we know that $\tan {{30}^{\circ }}=\dfrac{1}{\sqrt{3}}$ )

$\Rightarrow AC=3000\sqrt{3}$…………………………….. (2)

Now, we can clearly see in the figure $AC=AQ+QC$

i.e. $QC=AC-AQ$

from equation (1) and (2), we have

$AQ=3000m$ and

$AC=3000\sqrt{3}$ meters.

$\begin{align}

  & \therefore QC=\left( 3000\sqrt{3}-3000 \right)m \\

 & QC=3000\left( \sqrt{3}-1 \right)m \\

 & QC=3000\left( 1.732-1.000 \right)m \\

\end{align}$

(Since, we know that $\sqrt{3}=1.732$ )

$\begin{align}

  & \therefore QC=3000\left( 0.732 \right)m \\

 & QC=2196m \\

 & \Rightarrow PB=QC=2196m \\

\end{align}$

So, the speed of the aeroplane $=\dfrac{\text{Distance travelled}}{\text{time taken}}$ .

We have given time taken =15 seconds.

Distance travelled $PB=QC=2196m$

So, the speed of the aeroplane $=\dfrac{2196}{15}m/s$ .

$=146.6m/\sec $ .

Therefore, the speed of the aeroplane $=146.6m/\sec $.

 

Note: Students must note that in the question, we have calculated the speed of the aeroplane in unit meter per sec.

Student can also convert the speed on lm/hr given as-

We have speed of the aeroplane $=146.6m/\sec $. Speed of the aeroplane in km/hour.

$\begin{align}

  & \Rightarrow 146.4\times \dfrac{3600}{1000}km/hour \\

 & \Rightarrow 146.4\times \dfrac{18}{5}km/hour \\

 & \Rightarrow 527.04km/hour \\

\end{align}$

 


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