Solveeit Logo

Question

Question: An aeroplane when flying at a height \(2500m\) from the ground passes vertically above another aerop...

An aeroplane when flying at a height 2500m2500m from the ground passes vertically above another aeroplane. At an instant when the angles of elevation of the two aeroplanes from the same point on the ground are 450{45^0} and 300{30^0} respectively, then the vertical distance between the two aeroplanes at that instant is:
(a) 1158.5m(a){\text{ 1158}}{\text{.5m}}
(b) 1056.5m(b){\text{ 1056}}{\text{.5m}}
(c) 1008.5m(c){\text{ 1008}}{\text{.5m}}
(d) None of these(d){\text{ None of these}}

Explanation

Solution

Hint: In this question we have to find the vertical distance between the two aeroplanes so firstly draw the pictorial representation using the information provided in the question. Then in order to formulate the required equations which will be formed, use the trigonometric identities to find the answer.


Let us assume PQ=xPQ = x be the vertical height between the two aeroplanes.
And OQ=hOQ = h be the vertical height of the second aeroplane from the ground
Such that
x+h=2500x + h = 2500 … (1)
We are given that the angles of elevation of the two aeroplanes from the same point on the ground are 450{45^0} and 300{30^0}.
Now, let us first consider the triangle OAQOAQ, we have
tan300=OQ0A\tan {30^0} = \dfrac{{OQ}}{{0A}}
13=hl\Rightarrow \dfrac{1}{{\sqrt 3 }} = \dfrac{h}{l}
l=3h\Rightarrow l = \sqrt 3 h … (2)
Now, consider the triangle OAPOAP, we have
tan450=OP0A\tan {45^0} = \dfrac{{OP}}{{0A}}
x+hl=1\Rightarrow \dfrac{{x + h}}{l} = 1
x+h=l\Rightarrow x + h = l
Using equation (2), we get
x+h=3h\Rightarrow x + h = \sqrt 3 h …(3)
Now after equating the equation (1) and equation (3), we get
3h=2500\Rightarrow \sqrt 3 h = 2500
h=25003\Rightarrow h = \dfrac{{2500}}{{\sqrt 3 }}
h=1443.37m\Rightarrow h = 1443.37m
After substituting this value in equation (1), we get
x+1443.37=2500\Rightarrow x + 1443.37 = 2500
x=25001443.37\Rightarrow x = 2500 - 1443.37
x=1056.63m1056.5m\therefore x = 1056.63m \approx 1056.5m
Therefore, the vertical distance between the two aeroplanes at the given distance comes out to be 1056.5m1056.5m.
So, the required solution is option (b) 1056.5m(b){\text{ 1056}}{\text{.5m}}.

Note: Whenever we face such types of problems, the key point is to have a good understanding of the trigonometric functions like tan, cos, sin, sec etc. After putting the given values in the standard formula of these trigonometric ratios and then equating the results will help you reach the right answer for the problem.