Question

the angle of elevation of a jet fighter from a point of a on the ground is 60 degree after a flight of 15 seconds the angle of elevation changes to 30° if the jet is flying at the speed of 720 km per hour find the constant height at which the jet is flying​

Answer 1

Assumption:

Let A be the observation point and let B and C be the two positions of the jet fighter.

A.T.Q:

It takes 15 seconds to reach C from B. Let the height of jet fighter be BE=CD=h/km.

The speed of jet fighter is 720km/hr = 720000m/hr

As we know, Distance = speed × time

$⇒ BC = \frac{(720000 × 15)}{(60×60)} = 3000m = 3km$

In right-angled ∆BCA:-

$\implies{tan60 = \frac{BE}{AE}}$

$\implies{\sqrt{3} = \frac{h}{AE}}$

$∴ AE = \frac{h}{√3}$

In right-angled ∆CDA

$\implies{tan30 = \frac{CD}{AD}}$

$\implies{\frac{1}{√3} = \frac{h}{AD}}$

$\implies{AD = 4\sqrt{3}}$

$\implies{AD = AE + AD}$

$\implies{\sqrt{3h} = \frac{h}{√3} + 3}$

$\implies{\sqrt{3h} - \frac{h}{√3} = 3}$

$\implies{\frac{3h - h}{√3} = 3}$

$\implies{2h = 3\sqrt{3}}$

$\implies{h = \frac{3√3}{2}}$

$\implies{h = 3 × \frac{1.732}{2}}$

$\bold{\red{∴ h = 2.598km}}$

_____________________________