Question

The angle of elevation of helicopter from a point p on the ground is 45degree after 15 second the angle of elevation changes to 30°. If the helicopter is flying at a height of 4000m, find the speed of the helicopter (use1.7 for root 3)​

Answer 1

$\pink{\boxed{Question}}$

The angle of elevation of helicopter from a point P on the ground is 45° , after 15 seconds the angle of elevation changes to 30° . If the helicopter is flying at a height of 4000 m ,find the speed of helicopter. ( use 1.7 = √3 ).

$\pink{\boxed{Solution}}$

(Refer to the attachment for the figure firstly)

consider Δ ABP

tan ∠APB = AB / BP

tan 45° = AB / BP

 1 = AB / BP

$\large{\boxed{AB = BP=4000\:m}}$

now consider Δ DPC

tan ∠ DPC = DC / PC

tan 30° = DC / PC

1 / √3 = DC / PC

PC = DC √3

PC = 4000 × 1.7

$\large{\boxed{PC = 6800\:m}}$

Now,

BC = PC - BP = 6800 - 4000

$\large{\boxed{BC = 2800 \:m}}$

the time taken by helicopter to cover distance BC is 15 seconds so,

$speed\: of\: helicopter = \frac{distance\:BC}{time\:taken\:to\:cover\:BC}$

speed of helicopter= 2800 m / 15 sec

speed of helicopter= 560 / 3   ms⁻¹