Solution:

Consider A and D as the two positions of the aeroplane

AB is the height and P is the point

AB = 1 km

Take AD = x and PB = y

Angles of elevation from A and D at the point P are 60⁰ and 30⁰

Construct DC perpendicular to PB

DC = AB = 1 km

In right triangle APB

tan θ = AB/PB

Substituting the values

tan 60⁰ = 1/y

So we get

√3 = 1/y

y = 1/√3 ….. (1)

In right triangle DPC

tan θ = DC/PC

Substituting the values

tan 30⁰ = 1/ (x + y)

So we get

1/√3 = 1/ (x + y)

x + y = √3 ….. (2)

Using both the equations

x + 1/√3 = √3

By further calculation

x = √3 – 1/√3

x = (3 – 1)/ √3

x = 2/√3

Multiply and divide by √3

x = (2 × √3)/ (√3 × √3)

So we get

x = (2 × 1.732)/ 3

x = 3.464/3 km

This distance is covered in 10 seconds

Speed of aeroplane (in km/hr) = 3.464/3 × (60 × 60)/ 10

By further calculation

= 3464/ (3 × 1000) × 3600/10

So we get

= (3646 × 36)/ 300

= (3464 × 12)/ 100

= 41568/ 100

= 415.68 km/hr