Question

the angle of elevation of a jet fighter from point A on the ground is 60 degree. after a flight of 10sec the angle of elevation changes to 30 degree. if the jet is flying at speed of 648 km per hr,find the constant height at which jet is flying.​

Answer 1

Answer:

Step-by-step explanation:

The height is 15.5 m

Answer 2

Answer:

1.558 km

Step-by-step explanation:

The jet is flying at speed of 648 km per hr

Time = 10 sec = $\frac{10}{3600}hour$

Distance traveled in 10 sec = $Speed \times Time = 648 \times \frac{10}{3600}=1.8 km$

refer the attached figure

AB = height of of jet

BC = x

CD = 1.8 km

BD  x+1.8 km

In ΔABC

$Tan\theta = \frac{Perpendicular}{Base}\\Tan 60^{\circ}=\frac{AB}{BC}\\\sqrt{3}=\frac{AB}{x}\\\sqrt{3}x=AB$

In ΔABD

$Tan\theta = \frac{Perpendicular}{Base}\\Tan 30^{\circ}=\frac{AB}{BD}\\\frac{1}{\sqrt{3}}=\frac{AB}{x+1.8}\\\frac{1}{\sqrt{3}}(x+1.8)=AB$

So,$\frac{1}{\sqrt{3}}(x+1.8)=\sqrt{3}x$

$x+1.8=3x$

$1.8=2x$

$x=0.9$

$AB=\sqrt{3}x=\sqrt{3}(0.9)=1.558 km$

Hence  the constant height at which jet is flying.​ is 1.558 km