Mr. Vinod is a pilot in Air India. During the Covid-19 pandemic

Leave a Comment / Class 10th Case Study, trigonometry applications / By preeti

Q) 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° and at 10:01 am, it is 60°.

(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)
OR
Find the distance between passenger and airplane when the angle of elevation is 60º
(iii) Find the distance between passenger and airplane when the angle of elevation is 30°.

Ans:

(i) Distance covered by the airplane:

Given that the time taken to cover the distance d = 10:01 – 10:00 = 1 min

Also given the speed of the plane = 600 miles/hour = $\frac{600}{60}$ = 10 miles / min

Since we know that, the distance = Speed x time

∴ distance d = 10 x 1 = 10 miles

Therefore, the plane will cover 10 miles distance in the given time.

(ii) Altitude of the airplane:

It is given that the place is flying at the height of h, therefore we need to calculate the value of h.

from the diagram, we can deduce that, B’C’ = BC = d and AC’ = x

Similarly, BB’ = CC’ = h

Therefore. AB’ = AC’ + B’C’ = d + x

Let’s start from Δ ACC’,

tan 30° = $\frac{BB'}{AB'}$

$\frac{1}{\sqrt3} = \frac{h}{d + x}$

∴ h √3 = (d + x)…………. (i)

Next in Δ ACC’, tan 60° = $\frac{CC'}{AC'}$

∴ √3 = $\frac{h}{x}$

∴ x = $\frac{h}{\sqrt3}$ ………… (ii)

Substituting value of x from equation (ii) in equation (i), we get:

∴ h √3 = d + $\frac{h}{\sqrt3}$

∴ h (√3 – $\frac{1}{\sqrt3}$ ) = d = 10

∴ h (3 -1) = 10 √3

∴ h = 5√3 m

Therefore, height of the plane is 5√3 m.

OR

(ii) Distance between passenger and airplane at angle of elevation 60º:

Next in Δ ACC’, sin 60° = $\frac{CC'}{AC}$

∴ $\frac{\sqrt3}{2} = \frac{h}{AC}$

∴ AC = $\frac{2h}{\sqrt3}$

since h = 5 = $\frac{2 \times 5 \sqrt3}{\sqrt3}$ = 10 m

Therefore, distance between passengers and plane is 10 m, at elevation angle of 60⁰.

(iii) Distance between passenger and airplane at angle of elevation 30º:

From Δ ABB’, sin 30º = $\frac{CC'}{AB}$

$\frac{1}{2} = \frac{h}{AB}$

∴ AB = 2 h = 2 x 5√3 = 10√3 m

Therefore, distance between passengers and plane is 10√3 m, at elevation angle of 30⁰.

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