An aeroplane 3000 m high 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 60⁰ and 45⁰ respectively. Find the vertical distance between the two planes.

Solution:

Consider A and B as the two aeroplanes

P is a point on the ground such that 60⁰ and 45⁰ are the angles of elevations from A and B

AC = 3000 m

Take AC = 3000 m

BC = 3000 – x

PC = y

In right triangle APC

tan θ = AC/PC

Substituting the values

tan 60⁰ = 3000/y

So we get

√3 = 3000/y

y = 3000/√3 ….. (1)

In right triangle BPC

tan θ = BC/PC

Substituting the values

tan 45⁰ = (3000 – x)/ y

So we get

1 = (3000 – x)/ y

y = 3000 – x

Using equation (1)

3000/√3 = 3000 – x

By further calculation

x = 3000 – 3000/√3

Multiply and divide by √3

x = 3000 – (3000 × √3)/ (√3× √3)

So we get

x = 3000 – 1000 (1.732)

x = 3000 – 1732

x = 1268

Hence, the distance between the two planes is 1268 m.