The angle of elevation of a jet fighter from point A on ground is 60°. After flying 10 seconds, the angle change
30. If the jet is flying at a speed of 648 km/hour, find the constant height at which the jet is flying
Answers
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
3600
10
hour
Distance traveled in 10 sec = Speed \times Time = 648 \times \frac{10}{3600}=1.8 kmSpeed×Time=648×
3600
10
=1.8km
refer the attached figure
AB = height of of jet
BC = x
CD = 1.8 km
BD x+1.8 km
In ΔABC
\begin{lgathered}Tan\theta = \frac{Perpendicular}{Base}\\Tan 60^{\circ}=\frac{AB}{BC}\\\sqrt{3}=\frac{AB}{x}\\\sqrt{3}x=AB\end{lgathered}
Tanθ=
Base
Perpendicular
Tan60
∘
=
BC
AB
3
=
x
AB
3
x=AB
In ΔABD
\begin{lgathered}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\end{lgathered}
Tanθ=
Base
Perpendicular
Tan30
∘
=
BD
AB
3
1
=
x+1.8
AB
3
1
(x+1.8)=AB
So,\frac{1}{\sqrt{3}}(x+1.8)=\sqrt{3}x
3
1
(x+1.8)=
3
x
x+1.8=3xx+1.8=3x
1.8=2x1.8=2x
x=0.9x=0.9
AB=\sqrt{3}x=\sqrt{3}(0.9)=1.558 kmAB=
3
x=
3
(0.9)=1.558km
Hence the constant height at which jet is flying. is 1.558 km