Question
Question: An aeroplane when flying at a height of \( 3125m \) from the ground passes vertically below another ...
An aeroplane when flying at a height of 3125m from the ground passes vertically below another plane at an instant when the angles of elevation of the two planes from the same point on the ground are 30∘ and 60∘ respectively. Find the distance between the two planes at the instant.
Solution
Hint : Here, we will draw the diagram with the given specifications and with the help of trigonometric functions and the given angle we will find the unknown distance.
Complete step-by-step answer :
Let AB the surface of the ground, at point B there is the observer which finds two aeroplanes at point D and C with the angle of elevation of 30∘ and 60∘ .
We are given distance of one plane with the ground, AD=3125m
We have to find the distance between two planes, CD
In ΔDAB,
Tangent function is opposite upon the adjacent side.
⇒tan30∘=ABAD
Place the known values in the above equations,
31=AB3125
Do-cross multiplication
⇒AB=31253m ..... (a)
In ΔCBA ,
tan60∘=ABAD+DC
Place the known values in the above equations –
3=AB3125+DC
Do-cross multiplication and make AB the subject –
AB=33125+DC .... (b)
From the equations (a) and (b), we can equate each of the equation’s right hand side of the equation
31253=33125+DC
Do-cross multiplication and simplify
31253×3=3125+DC
Apply the property of the product of two same square-roots given the number.
3125×3=3125+DC 9375=3125+DC
When any term changes its sides, the sign also changes. Positive becomes negative and negative becomes positive.
⇒DC=9375−3125 ⇒DC=6250m
Hence, the distance between the two planes is 6250m.
Note : Remember the basic difference between the angles of the elevation and the depression and apply accordingly. And remember the basic trigonometric functions and its values for direct substitution of its correct value.