Question

An aeroplane flying horizontally 1km above the ground is observed at an elevation of ${60^\circ }$ After 10 seconds, its elevation is observed to be \[{30^\circ }\]. Find the speed of the aeroplane in kmph.

Answer 1

Hint: start the solution by listing down the given information and constructing the diagram accordingly. Then try to find the distance covered by the aeroplane first and then you will get the distance covered easily. Finally the answer asked to us is in kmph, hence try to do the necessary changes in the final answer.

Complete step by step answer:
let’s start the solution by listing down what is given to us and then constructing a diagram for our reference.
Initially the plane flying above 1Km was observed at an elevation of ${60^\circ }$
After 10 seconds the same plane at the same height was observed at an elevation of \[{30^\circ }\].

Here, O is the point of observation when the plane was observed at an angle of ${60^\circ }$
Let P be the position of the plane. Hence $\angle POA = {60^\circ }$ and the distance $PA = 1km$
Now after 10 seconds the position of the plane is at Q at the same height of 1km. hence the distance $QB = 1km$. Also the elevation is observed to be \[{30^\circ }\]. Hence $\angle QOB = {30^\circ }$
First we will try to find the distance OA and OB so that we can calculate the distance AB
Consider the right angle triangle $\Delta POA$.
$\tan \theta = \dfrac{{opposite}}{{adjacent}}$
$ \Rightarrow \tan {60^\circ } = \dfrac{{PA}}{{OA}}$
since $\tan {60^\circ } = \sqrt 3 $, AC=1 substituting we get
$
  \sqrt 3 = \dfrac{1}{{OA}} \\
   \Rightarrow OA = \dfrac{1}{{\sqrt 3 }} \\
 $
Also, in right angled triangle QOB,
$
  \tan {30^\circ } = \dfrac{{QB}}{{OB}} \\
   \Rightarrow \dfrac{1}{{\sqrt 3 }} = \dfrac{1}{{OB}} \\
   \Rightarrow OB = \sqrt 3 \\
 $
Now, AB= OB-OA
${\text{AB = }}\sqrt 3 - \dfrac{1}{{\sqrt 3 }}$
$ = \dfrac{{3 - 1}}{{\sqrt 3 }}$
$ = \dfrac{2}{{\sqrt 3 }}$
Hence from this we observe that the distance covered by the aeroplane in 10 seconds will be $\dfrac{2}{{\sqrt 3 }}$km
Also since we have been told to find the answer in kmph, we will change the time from seconds to hour.
Hence, time = 10seconds=$\dfrac{{10}}{{3600}}hours = \dfrac{1}{{360}}hours$
To find the speed of the aeroplane we know the formula,
${\text{speed = }}\dfrac{{{\text{distance}}}}{{{\text{time}}}}$
$ \Rightarrow \dfrac{{\dfrac{2}{{\sqrt 3 }}}}{{\dfrac{1}{{360}}}} = \dfrac{{360 \times 2}}{{\sqrt 3 }} = \dfrac{{720}}{{1.732}} = 415.68$
hence the speed of the aeroplane will be 415.68kmph

Note: this type of questions get easier or are easy to understand if we represent the given information in graphical form. Also keep attention to the units asked in the question as this can make a huge difference in the answer. Always mention the units of the answer obtained as this brings sense to the answer.