Question 14 Marks
Read the following text carefully and answer the questions that follow:
Mr. Vinod is a pilot in Air India. During the Covid $-19$ pandemic, many Indian passengers were stuck at Dubai Airport. The government of India sent special aircraft to take them. Mr. Vinod was leading this operation. He is flying from Dubai to New Delhi with these passengers. His airplane is approaching point $A$ along a straight line and at a constant altitude h. At $10:00 \ am$, the angle of elevation of the airplane is $30^\circ $ and at $10:01 am$, it is $60^\circ .$
$i.$ What is the distance d is covered by the airplane from $10:00 \ am$ to $10:01 \ am$ if the speed of the airplane is constant and equal to $600$ miles/hour?
$ii.$ What is the altitude h of the airplane? $($round answer to $2$ decimal places$)$
$iii.$ Find the distance between passenger and airplane when the angle of elevation is $30^\circ .$
OR
Find the distance between passenger and airplane when the angle of elevation is $60^\circ .$
Mr. Vinod is a pilot in Air India. During the Covid $-19$ pandemic, many Indian passengers were stuck at Dubai Airport. The government of India sent special aircraft to take them. Mr. Vinod was leading this operation. He is flying from Dubai to New Delhi with these passengers. His airplane is approaching point $A$ along a straight line and at a constant altitude h. At $10:00 \ am$, the angle of elevation of the airplane is $30^\circ $ and at $10:01 am$, it is $60^\circ .$
$i.$ What is the distance d is covered by the airplane from $10:00 \ am$ to $10:01 \ am$ if the speed of the airplane is constant and equal to $600$ miles/hour?
$ii.$ What is the altitude h of the airplane? $($round answer to $2$ decimal places$)$
$iii.$ Find the distance between passenger and airplane when the angle of elevation is $30^\circ .$
OR
Find the distance between passenger and airplane when the angle of elevation is $60^\circ .$
Answer
View full question & answer→$i.$ Time covered $10.00 \ am$ to $10.01 am =1$ minute $=\frac{1}{60}$ hour
Given: Speed $=600$ miles hour
Thus, distance $d =600 \times \frac{1}{60}=10$ miles
$ii.$ Now, $\tan 30^{\circ}=\frac{B B^{\prime}}{B^{\prime} A}=\frac{h}{10+x}..$
And $\tan 60^{\circ}=\frac{C C^{\prime}}{C^{\prime} A}=\frac{B B^{\prime}}{C^{\prime} A}=\frac{h}{x}$
$x=\frac{h}{\tan 60^{\circ}}=\frac{h}{\sqrt{3}}$
Putting the value of $x$ in eq$(1),$ we get,
$\tan 30^{\circ}=\frac{h}{10+\frac{h}{\sqrt{3}}}=\frac{\sqrt{3} h}{10 \sqrt{3}+h}$
$\tan 30^0=\frac{\sqrt{3} h}{10 \sqrt{3}+h}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{\sqrt{3} h}{10 \sqrt{3}+h}$
$\Rightarrow 3 h=10 \sqrt{3}+h$
$\Rightarrow 2 h=10 \sqrt{3}$
$\Rightarrow h=5 \sqrt{3}=8.66 \text { miles }$
Thus, the altitude $'h\ '$ of the airplane is $8.66$ miles.
$iii.$ The distance between passenger and airplane when the angle of elevation is $30^{\circ}$.
$\text { In } \triangle ABB^{\prime}$
$\sin 30^{\circ}=\frac{B B^{\prime}}{A B}$
$\Rightarrow \frac{1}{2}=\frac{8.66}{A B}$
$\Rightarrow AB=17.32 \text { miles }$
OR
The distance between passenger and airplane when the angle of elevation is $60^{\circ}$.
$\text { In } \triangle ACC^{\prime}$
$\sin 60^{\circ}=\frac{C C^{\prime}}{A C}$
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{5 \sqrt{3}}{A C}$
$\Rightarrow AC=10 \text { miles }$
Given: Speed $=600$ miles hour
Thus, distance $d =600 \times \frac{1}{60}=10$ miles
$ii.$ Now, $\tan 30^{\circ}=\frac{B B^{\prime}}{B^{\prime} A}=\frac{h}{10+x}..$
And $\tan 60^{\circ}=\frac{C C^{\prime}}{C^{\prime} A}=\frac{B B^{\prime}}{C^{\prime} A}=\frac{h}{x}$
$x=\frac{h}{\tan 60^{\circ}}=\frac{h}{\sqrt{3}}$
Putting the value of $x$ in eq$(1),$ we get,
$\tan 30^{\circ}=\frac{h}{10+\frac{h}{\sqrt{3}}}=\frac{\sqrt{3} h}{10 \sqrt{3}+h}$
$\tan 30^0=\frac{\sqrt{3} h}{10 \sqrt{3}+h}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{\sqrt{3} h}{10 \sqrt{3}+h}$
$\Rightarrow 3 h=10 \sqrt{3}+h$
$\Rightarrow 2 h=10 \sqrt{3}$
$\Rightarrow h=5 \sqrt{3}=8.66 \text { miles }$
Thus, the altitude $'h\ '$ of the airplane is $8.66$ miles.
$iii.$ The distance between passenger and airplane when the angle of elevation is $30^{\circ}$.
$\text { In } \triangle ABB^{\prime}$
$\sin 30^{\circ}=\frac{B B^{\prime}}{A B}$
$\Rightarrow \frac{1}{2}=\frac{8.66}{A B}$
$\Rightarrow AB=17.32 \text { miles }$
OR
The distance between passenger and airplane when the angle of elevation is $60^{\circ}$.
$\text { In } \triangle ACC^{\prime}$
$\sin 60^{\circ}=\frac{C C^{\prime}}{A C}$
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{5 \sqrt{3}}{A C}$
$\Rightarrow AC=10 \text { miles }$
