Question

The angle of elevation of a jet fighter from a point A on the ground is 60°. After a flight of 15 seconds, the angle of elevation change to 30°. If the jet is flying at s speed of 720km/h, find the constant height at which the jet is flying. (proper solution with language, diagram etc.)

Answer 1

Distance travelled in 1 hour (3600 seconds) = 720 km.
Distance travelled in 1 second
$= \frac{720}{3600} = \frac{1}{5} \: km = \frac{1000}{5} m = 200 \: m$
Distance travelled in 15 seconds
$= 200 \times 15 = 3000 \: m$
AE = CD = 3000 m
AC = ED
In triangle ABC,
$\tan(60) = \frac{ac}{bc} \\ \sqrt{3} = \frac{ac}{bc} \\ bc = \frac{ac}{ \sqrt{3} }$
In triangle BED,
$\tan(30) = \frac{ed}{bd}$
$\frac{1}{ \sqrt{3} } = \frac{ac}{bc + cd}$
$\frac{1}{ \sqrt{3} } = \frac{ac}{ \frac{ac}{ \sqrt{3} } + 3000 }$
$\frac{1}{ \sqrt{3} } = \frac{ac}{ \frac{ac + 3000 \sqrt{3} }{ \sqrt{3} } }$
$\frac{1}{ \sqrt{3} } = ac \times \frac{ \sqrt{3} }{ac + 3000 \sqrt{3} }$
$ac = \frac{1}{ \sqrt{3} } \times \frac{ac + 3000 \sqrt{3} }{ \sqrt{3} } = \frac{ac + 3000 \sqrt{3} }{3}$
$3ac = ac + 3000 \sqrt{3}$
$3ac - ac = 3000 \sqrt{ 3 }$
$2ac = 3000 \sqrt{3}$
$ac = \frac{3000 \sqrt{3} }{2} = 1500 \sqrt{3} \: m$
Hence, the constant height at which jet plane is flying
$= 1500 \sqrt{3} \: m$
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