The jet is flying at speed of 648 km per hr
Time = 10 sec = \(\frac {10}{100}\) hour
Distance traveled in 10 sec = Speed x Time
\(= 648 \times \frac {10}{3600}\)
\(= 1.8 km\)
AB = height of of jet
BC = x
CD = 1.8 km
BD = x + 1.8 km
In ΔABC,
\(\tan \theta = \frac{\text{Perpendicular}}{\text{Base}}\)
\(\tan 60°= \frac{AB}{BC}\)
\(\sqrt3= \frac{AB}{x}\)
\(\sqrt3 x= {AB}\)
In ΔABD
\(\tan \theta = \frac{\text{Perpendicular}}{\text{Base}}\)
\(\tan 30°= \frac{AB}{BD}\)
\(\frac 1{\sqrt 3} = \frac{AB}{x+1.8}\)
\(\frac 1{\sqrt 3}{(x+1.8)} = {AB}\)
So, \(\frac 1{\sqrt 3}{(x+1.8)} = \sqrt 3x\)
\(x + 1.8 = 3x\)
\(1.8 =2x\)
\(x = 0.9\)
\(AB = \sqrt 3x\)
\(= \sqrt3(0.9)\)
\( = 1.558 km\)
Hence, the constant height at which jet is flying is 1.558 km.
