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